NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT solutions for class 10 maths chapter 6 Triangles: We have studied about triangles and properties of the triangles in the previous classes. Here we are going to study similar triangles and their properties. Solutions of NCERT for class 10 maths chapter 6 Triangles is covering a detailed explanation to each and every question available in the practice exercises. This chapter starts by introducing the difference between similar and congruent figures. If two figures are the same in shape and size, the figures are said to be congruent. Two figures that have the same shape but not necessarily the same size are called similar figures. For example, all circles are similar but only those circles with the same radii are congruent which means all congruent figures are similar but converse may not be true. In this chapter of CBSE NCERT solutions for class 10 maths chapter 6 Triangles, you will get the solutions to the problems based on the similarity of triangles. The concept of similarity of triangles can be used to measure the height of objects. Below is a question solved in NCERT solutions for class 10 maths chapter 6 Triangles. This question can be solved using the concept of similarity of triangles.

Question: A girl of height 90 cm is walking away from the base of a lamp-post at a speed of 1.2 m/s. If the lamp is 3.6 m above the ground, find the length of her shadow after 4 seconds.

Given:-

length of lamp post (AB) = 3.6m

Height of the girl = 90cm or 0.9m

Speed of the girl = 1.2m/sec

To find:- the length of the shadow DE.

Solution:- the girl walked BD distance in 4 seconds.

The distance traveled girl = length of BD = 4 x 1.2 => 4.8m

In Triangle ABE and Triangle CDE

$\angle A = \angle A (Common)$

$\angle B = \angle D (90 degree)$

Thus Triangle ABE and Triangle CDE are similar triangles.

From the theorem:- If two triangles are similar then the ratio of their sides is equal.
$\frac{BE}{DE} = \frac{AB}{CD}$

$\frac{BD+DE}{DE} = \frac{AB}{CD}$

$\frac{4.8+DE}{DE} = \frac{3.6}{0.9}$

$DE= 1.6m$

Hence the length of the shadow is 1.6m.

These types of examples are mentioned in this chapter. Class wise and chapter-wise NCERT solutions can be downloaded by clicking on the above-given link.

Types of questions asked from class 10 maths chapter 6 Triangles

• Similarity of triangles

• Theorems based on similar triangles

• Areas of similar triangles

• Theorems related to Trapezium

• Pythagoras theorem

CBSE NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6.1

All circles are similar.

Since all the circles have a similar shape. They may have different radius but the shape of all circles is the same.

Therefore, all circles are similar.

All squares are similar.

Since all the squares have a similar shape. They may have a different side but the shape of all square is the same.

Therefore, all squares are similar.

All equilateral triangles are similar.

Since all the equilateral triangles have a similar shape. They may have different sides but the shape of all equilateral triangles is the same.

Therefore, all equilateral triangles are similar.

Two polygons of the same number of sides are similar if their corresponding angles are equal and their corresponding sides are proportional.

Thus, (a) equal

(b) proportional

The  two different examples of a pair of  similar figures are :

1. Two circles with different radii.

2. Two rectangles with different breadth and length.

The  two different examples of a pair of  non-similar figures are :

1.Rectangle and circle

2. A circle and a triangle.

Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional  i.e. $1:2$  but their corresponding angles are not equal.

NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6.2

(i)

Let EC be x

Given: DE || BC

By using the proportionality theorem, we get

$\frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow \frac{1.5}{3}=\frac{1}{x}$

$\Rightarrow x=\frac{3}{1.5}=2\, cm$

$\therefore EC=2\, cm$

(ii)

Given: DE || BC

By using the proportionality theorem, we get

$\frac{AD}{DB}=\frac{AE}{EC}$

$\Rightarrow \frac{x}{7.2}=\frac{1.8}{5.4}$

$\Rightarrow x=\frac{7.2}{3}=2.4\, cm$

$\therefore AD=2.4\, cm$

(i)

Given :

PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

$\frac{PE}{EQ}=\frac{3.9}{3}=1.3\, cm$                       and               $\frac{PF}{FR}=\frac{3.6}{2.4}=1.5\, cm$

We have

$\frac{PE}{EQ} \neq \frac{PF}{FR}$

Hence, EF is not parallel to QR.

(ii)

Given :

PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

$\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}\, cm$                       and               $\frac{PF}{FR}=\frac{8}{9}\, cm$

We have

$\frac{PE}{EQ} = \frac{PF}{FR}$

Hence, EF is parallel to QR.

(iii)

Given :

PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

$\frac{PE}{PQ}=\frac{0.18}{1.28}=\frac{9}{64}\, cm$                       and               $\frac{PF}{PR}=\frac{0.36}{2.56}=\frac{9}{64}\, cm$

We have

$\frac{PE}{EQ} = \frac{PF}{FR}$

Hence, EF is parallel to QR.

Given : LM || CB and LN || CD

To prove :

$\frac{AM}{AB} = \frac{AN}{AD }$

Since , LM || CB so we have

$\frac{AM}{AB}=\frac{AL}{AC}.............................................1$

Also, LN || CD

$\frac{AL}{AC}=\frac{AN}{AD}.............................................2$

From equation 1 and 2, we have

$\frac{AM}{AB} = \frac{AN}{AD }$

Hence proved.

Given :  DE || AC and DF || AE.

To prove :

$\frac{BF}{FE} = \frac{BE}{EC }$

Since , DE || AC so we have

$\frac{BD}{DA}=\frac{BE}{EC}.............................................1$

Also,DF || AE

$\frac{BD}{DA}=\frac{BF}{FE}.............................................2$

From equation 1 and 2, we have

$\frac{BF}{FE} = \frac{BE}{EC }$

Hence proved.

Given : DE || OQ  and DF || OR.

To prove EF || QR.

Since DE || OQ so we have

$\frac{PE}{EQ}=\frac{PD}{DO}.............................................1$

Also, DF || OR

$\frac{PF}{FR}=\frac{PD}{DO}.............................................2$

From equation 1 and 2, we have

$\frac{PE}{EQ} = \frac{PF}{FR }$

Thus,  EF || QR.    (converse of basic proportionality theorem)

Hence proved.

Given : AB || PQ and AC || PR

To prove: BC || QR

Since,  AB || PQ so we have

$\frac{OA}{AP}=\frac{OB}{BQ}.............................................1$

Also, AC || PR

$\frac{OA}{AP}=\frac{OC}{CR}.............................................2$

From equation 1 and 2, we have

$\frac{OB}{BQ} = \frac{OC}{CR }$

Therefore, BC || QR. (converse basic proportionality theorem)

Hence proved.

Let PQ is a line passing through the midpoint of line AB and parallel to line BC intersecting line AC at point Q.

i.e.$PQ||BC$   and   $AP=PB$.

Using basic proportionality theorem, we have

$\frac{AP}{PB}=\frac{AQ}{QC}..........................1$

Since $AP=PB$

$\frac{AQ}{QC}=\frac{1}{1}$

$\Rightarrow AQ=QC$

$\therefore$ Q is the midpoint of AC.

Let P is the midpoint of line AB and Q is the midpoint of line AC.

PQ is the line joining midpoints P and Q of line AB and AC, respectively.

i.e.$AQ=QC$   and   $AP=PB$.

we have,

$\frac{AP}{PB}=\frac{1}{1}..........................1$

$\frac{AQ}{QC}=\frac{1}{1}...................................2$

From equation 1 and 2, we get

$\frac{AQ}{QC}=\frac{AP}{PB}$

$\therefore$ By basic proportionality theorem, we have $PQ||BC$

Draw a line EF passing through point O such that $EO||CD\, \, and\, \, FO||CD$

To prove :

$\frac{AO}{BO} = \frac{CO}{DO}$

In $\triangle ADC$, we have $CD||EO$

So, by using basic proportionality theorem,

$\frac{AE}{ED}=\frac{AO}{OC}........................................1$

In $\triangle ABD$, we have $AB||EO$

So, by using basic proportionality theorem,

$\frac{DE}{EA}=\frac{OD}{BO}........................................2$

Using equation 1 and 2, we get

$\frac{AO}{OC}=\frac{BO}{OD}$

$\Rightarrow \frac{AO}{BO} = \frac{CO}{DO}$

Hence proved.

Draw a line EF passing through point O such that $EO||AB$

Given  :

$\frac{AO}{BO} = \frac{CO}{DO}$

In $\triangle ABD$, we have $AB||EO$

So, by using basic proportionality theorem,

$\frac{AE}{ED}=\frac{BO}{DO}........................................1$

However, its is given that

$\frac{AO}{CO} = \frac{BO}{DO}..............................2$

Using equation 1 and 2 , we get

$\frac{AE}{ED}=\frac{AO}{CO}$

$\Rightarrow EO||CD$         (By basic proportionality theorem)

$\Rightarrow AB||EO||CD$

$\Rightarrow AB||CD$

Therefore, ABCD is a trapezium.

NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6.3

(i) $\angle A=\angle P=60 \degree$

$\angle B=\angle Q=80 \degree$

$\angle C=\angle R=40 \degree$

$\therefore \triangle ABC \sim \triangle PQR$     (By AAA)

So , $\frac{AB}{QR}=\frac{BC}{RP}=\frac{CA}{PQ}$

(ii) As corresponding sides of both triangles are proportional.

$\therefore \triangle ABC \sim \triangle PQR$    (By SSS)

(iii) Given triangles are not similar because corresponding sides are not proportional.

(iv) $\triangle MNL \sim \triangle PQR$  by SAS similarity criteria.

(v) Given triangles are not similar because the corresponding angle is not contained by two corresponding sides

(vi) In $\triangle DEF$, we know that

$\angle D+\angle E+\angle F=180 \degree$

$\Rightarrow 70 \degree+80 \degree+\angle F=180 \degree$

$\Rightarrow 150 \degree+\angle F=180 \degree$

$\Rightarrow \angle F=180 \degree-150 \degree=30 \degree$

In $\triangle PQR$, we know that

$\angle P+\angle Q+\angle R=180 \degree$

$\Rightarrow 30 \degree+80 \degree+\angle R=180 \degree$

$\Rightarrow 110 \degree+\angle R=180 \degree$

$\Rightarrow \angle R=180 \degree-110 \degree=70 \degree$

$\angle Q=\angle P=70 \degree$

$\angle E=\angle Q=80 \degree$

$\angle F=\angle R=30 \degree$

$\therefore \triangle DEF\sim \triangle PQR$   ( By AAA)

Given :  $\Delta ODC \sim \Delta OBA$, $\angle BOC = 125 \degree$  and $\angle CDO = 70 \degree$

$\angle DOC+\angle BOC=180 \degree$            (DOB is a straight line)

$\Rightarrow \angle DOC+125 \degree=180 \degree$

$\Rightarrow \angle DOC=180 \degree-125 \degree$

$\Rightarrow \angle DOC=55 \degree$

In $\Delta ODC ,$

$\angle DOC+\angle ODC+\angle DCO=180 \degree$

$\Rightarrow 55 \degree+ 70 \degree+\angle DCO=180 \degree$

$\Rightarrow \angle DCO+125 \degree=180 \degree$

$\Rightarrow \angle DCO=180 \degree-125 \degree$

$\Rightarrow \angle DCO=55 \degree$

Since ,$\Delta ODC \sim \Delta OBA$ , so

$\Rightarrow\angle OAB= \angle DCO=55 \degree$   ( Corresponding angles are equal in similar triangles).

In  $\triangle DOC\, and\, \triangle BOA$ , we have

$\angle CDO=\angle ABO$        ( Alternate interior angles as $AB||CD$)

$\angle DCO=\angle BAO$        ( Alternate interior angles as $AB||CD$)

$\angle DOC=\angle BOA$        ( Vertically opposite angles are equal)

$\therefore \triangle DOC\, \sim \, \triangle BOA$    ( By AAA)

$\therefore \frac{DO}{BO}=\frac{OC}{OA}$      ( corresponding sides are equal)

$\Rightarrow \frac{OA }{OC} = \frac{OB }{OD }$

Hence proved.

Given :  $\frac{QR }{QS } = \frac{QT}{PR}$  and $\angle 1 = \angle 2$

To prove : $\Delta PQS \sim \Delta TQR$

In $\triangle PQR$ , $\angle PQR=\angle PRQ$

$\therefore PQ=PR$

$\frac{QR }{QS } = \frac{QT}{PR}$                 (Given)

$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$

In $\Delta PQS\, and\, \Delta TQR$,

$\Rightarrow \frac{QR }{QS } = \frac{QT}{PQ}$

$\angle Q=\angle Q$        (Common)

$\Delta PQS \sim \Delta TQR$   ( By SAS)

Given : $\angle$ P = $\angle$ RTS

To prove RPQ ~ $\Delta$ RTS.

In  $\Delta$ RPQ  and  $\Delta$ RTS,

$\angle$ P = $\angle$ RTS    (Given )

$\angle$ R = $\angle$ R         (common)

$\Delta$ RPQ ~ $\Delta$ RTS.  ( By AA)

Given : $\triangle ABE \cong \triangle ACD$

To prove ADE ~ $\Delta$ ABC.

Since $\triangle ABE \cong \triangle ACD$

$AB=AC$          ( By CPCT)

$AD=AE$          (By CPCT)

In $\Delta$ADE  and  $\Delta$ ABC,

$\angle A=\angle A$    ( Common)

and

$\frac{AD}{AB}=\frac{AE}{AC}$       (  $AB=AC$   and   $AD=AE$ )

Therefore, $\Delta$ADE ~ $\Delta$ ABC.   ( By SAS criteria)

To prove : $\Delta AEP \sim \Delta CDP$

In $\Delta AEP \, \, and\, \, \Delta CDP$,

$\angle AEP=\angle CDP$     ( Both angles are right angle)

$\angle APE=\angle CPD$        (Vertically opposite angles )

$\Delta AEP \sim \Delta CDP$    ( By AA criterion)

To prove : $\Delta ABD \sim \Delta CBE$

In $\Delta ABD \, \, and\, \, \Delta CBE$,

$\angle ADB=\angle CEB$     ( Both angles are right angle)

$\angle ABD=\angle CBE$        (Common )

$\Delta ABD \sim \Delta CBE$    ( By AA criterion)

To prove : $\Delta AEP \sim \Delta ADB$

In $\Delta AEP \, \, \, and\, \, \Delta ADB$,

$\angle AEP=\angle ADB$     ( Both angles are right angle)

$\angle A=\angle A$        (Common )

$\Delta AEP \sim \Delta ADB$    ( By AA criterion)

To prove : $\Delta PDC \sim \Delta BEC$

In $\Delta PDC \, \, and\, \, \, \Delta BEC$,

$\angle CDP=\angle CEB$     ( Both angles are right angle)

$\angle C=\angle C$        (Common )

$\Delta PDC \sim \Delta BEC$    ( By AA criterion)

To prove : $\Delta ABE \sim \Delta CFB$

In $\Delta ABE \, \, \, and\, \, \Delta CFB$,

$\angle A=\angle C$   ( Opposite angles of a parallelogram are equal)

$\angle AEB=\angle CBF$  ( Alternate angles of AE||BC)

$\Delta ABE \sim \Delta CFB$     ( By AA criterion )

To prove :  $\Delta ABC \sim \Delta AMP$

In $\Delta ABC \, \, and\, \, \Delta AMP$,

$\angle ABC=\angle AMP$     ( Each $90 \degree$)

$\angle A=\angle A$                     ( common)

$\Delta ABC \sim \Delta AMP$   ( By AA criterion )

To prove :

$\frac{CA }{PA } = \frac{BC }{MP}$

In $\Delta ABC \, \, and\, \, \Delta AMP$,

$\angle ABC=\angle AMP$     ( Each $90 \degree$)

$\angle A=\angle A$                     ( common)

$\Delta ABC \sim \Delta AMP$   ( By AA criterion )

$\frac{CA }{PA } = \frac{BC }{MP}$                ( corresponding parts of similar triangles )

Hence proved.

To prove :

$\frac{CD}{GH} = \frac{AC}{FG}$

Given : $\Delta ABC \sim \Delta EGF$

$\angle A=\angle F,\angle B=\angle E\, \, and \, \, \angle ACB=\angle FGE,\angle ACB=\angle FGE$

$\therefore \angle ACD=\angle FGH$    ( CD and GH are bisectors of equal angles)

$\therefore \angle DCB=\angle HGE$   ( CD and GH are bisectors of equal angles)

In $\Delta ACD \, \, and\, \, \Delta FGH$

$\therefore \angle ACD=\angle FGH$     ( proved above)

$\angle A=\angle F$                        ( proved above)

$\Delta ACD \sim \Delta FGH$       ( By AA criterion)

$\Rightarrow \frac{CD}{GH} = \frac{AC}{FG}$

Hence proved.

To prove : $\Delta DCB \sim \Delta HGE$

Given : $\Delta ABC \sim \Delta EGF$

In $\Delta DCB \,\, \, and\, \, \Delta HGE$,

$\therefore \angle DCB=\angle HGE$   ( CD and GH are bisectors of equal angles)

$\angle B=\angle E$           ( $\Delta ABC \sim \Delta EGF$)

$\Delta DCB \sim \Delta HGE$     ( By AA criterion )

To prove : $\Delta DCA \sim \Delta HGF$

Given : $\Delta ABC \sim \Delta EGF$

In $\Delta DCA \, \, \, and\, \, \Delta HGF$,

$\therefore \angle ACD=\angle FGH$   ( CD and GH are bisectors of equal angles)

$\angle A=\angle F$           ( $\Delta ABC \sim \Delta EGF$)

$\Delta DCA \sim \Delta HGF$     ( By AA criterion )

To prove : $\Delta ABD \sim \Delta ECF$

Given: ABC is an isosceles triangle.

$AB=AC \, \, and\, \, \angle B=\angle C$

In $\Delta ABD \, \, and\, \, \Delta ECF$,

$\angle ABD=\angle ECF$         ($\angle ABD=\angle B=\angle C=\angle ECF$)

$\angle ADB=\angle EFC$        ( Each $90 \degree$)

$\Delta ABD \sim \Delta ECF$          ( By AA criterion)

AD and PM are medians of triangles. So,

$BD=\frac{BC}{2}\, and\, QM=\frac{QR}{2}$

Given :

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AD}{PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{\frac{1}{2}BC}{\frac{1}{2}QR}=\frac{AD}{PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

In $\triangle ABD\, and\, \triangle PQM,$

$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

$\therefore \triangle ABD\sim \triangle PQM,$         (SSS similarity)

$\Rightarrow \angle ABD=\angle PQM$      ( Corresponding angles of similar triangles )

In $\triangle ABC\, and\, \triangle PQR,$

$\Rightarrow \angle ABD=\angle PQM$    (proved above)

$\frac{AB}{PQ}=\frac{BC}{QR}$

Therefore,$\Delta ABC \sim \Delta PQR$.   ( SAS similarity)

In, $\triangle ADC \, \, and\, \, \triangle BAC,$

$\angle ADC = \angle BAC$             ( given )

$\angle ACD = \angle BCA$              (common )

$\triangle ADC \, \, \sim \, \, \triangle BAC,$      ( By AA rule)

$\frac{CA}{CB}=\frac{CD}{CA}$    ( corresponding sides of similar triangles )

$\Rightarrow CA^2=CB\times CD$

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$              (given)

Produce AD and PM to E and L such that AD=DE  and PM=DE. Now,

join B to E,C to E,Q to L and R to L.

AD and PM are medians of a triangle, therefore

QM=MR  and    BD=DC

PM=ML               (By construction)

So, diagonals of ABEC bisecting each other at D,so ABEC is a parallelogram.

Similarly, PQLR is also a parallelogram.

Therefore, AC=BE ,AB=EC and PR=QL,PQ=LR

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{AD}{PM}$               (Given )

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{2.AD}{2.PM}$

$\Rightarrow \frac{AB}{PQ}=\frac{BE}{QL}=\frac{AE}{PL}$

$\Delta ABE \sim \Delta PQL$                (SSS similarity)

$\angle BAE=\angle QPL$ ...................1           (Corresponding angles of similar triangles)

Similarity, $\triangle AEC=\triangle PLR$

$\angle CAE=\angle RPL$........................2

$\angle BAE+\angle CAE=\angle QPL+\angle RPL$

$\angle CAB=\angle RPQ$............................3

In $\triangle ABC\, and\, \, \triangle PQR,$

$\frac{AB}{PQ}=\frac{AC}{PR}$        ( Given )

$\angle CAB=\angle RPQ$        ( From above equation 3)

$\triangle ABC\sim \triangle PQR$         ( SAS similarity)

CD = pole

AB = tower

In $\triangle ABE\, \, and\, \triangle CDF,$

$\angle CDF=\angle ABE$            ( Each $90 \degree$)

$\angle DCF=\angle BAE$             (Angle of sun at same place )

$\triangle ABE\, \, \sim \, \triangle CDF,$          (AA similarity)

$\frac{AB}{CD}=\frac{BE}{QL}$

$\Rightarrow \frac{AB}{6}=\frac{28}{4}$

$\Rightarrow AB=42$  cm

Hence, the height of the tower is 42 cm.

$\Delta AB C \sim \Delta PQR$     ( Given )

$\frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}$     ............... ....1( corresponding sides of similar triangles )

$\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R$....................................2

AD and PM are medians of triangle.So,

$BD=\frac{BC}{2}\, and\, QM=\frac{QR}{2}$   ..........................................3

From equation 1 and 3, we have

$\frac{AB}{PQ}=\frac{BD}{QM}$...................................................................4

In $\triangle ABD\, and\, \triangle PQM,$

$\angle B=\angle Q$        (From equation 2)

$\frac{AB}{PQ}=\frac{BD}{QM}$        (From equation 4)

$\triangle ABD\, \sim \, \triangle PQM,$        (SAS similarity)

$\frac{AB}{PQ}=\frac{BD}{QM}=\frac{AD}{PM}$

NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6

$\Delta ABC \sim \Delta DEF$              ( Given )

ar(ABC) = 64 $cm^2$ and ar(DEF)=121 $cm^2$.

EF = 15.4 cm                              (Given )

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}$

$\frac{64}{121}=\frac{BC^2}{(15.4)^2}$

$\Rightarrow \frac{8}{11}=\frac{BC}{15.4}$

$\Rightarrow \frac{8\times 15.4}{11}=BC$

$\Rightarrow BC=11.2 cm$

Given: Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O.

AB = 2 CD          ( Given )

In $\triangle AOB\, and\, \triangle COD,$

$\angle COD=\angle AOB$                (vertically opposite angles )

$\angle OCD=\angle OAB$                 (Alternate angles)

$\angle ODC=\angle OBA$                 (Alternate angles)

$\therefore \triangle AOB\, \sim \, \triangle COD$               (AAA similarity)

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{AB^2}{CD^2}$

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{(2CD)^2}{CD^2}$

$\frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4.CD^2}{CD^2}$

$\Rightarrow \frac{ar(\triangle AOB)}{ar(\triangle COD)}=\frac{4}{1}$

$\Rightarrow ar(\triangle AOB)=ar(\triangle COD)=4:1$

Let DM and AP be perpendicular on BC.

$area\,\,of\,\,triangle=\frac{1}{2}\times base\times perpendicular$

$\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}$

In $\triangle APO\, and\, \triangle DMO,$

$\angle APO=\angle DMO$           (Each $90 \degree$)

$\angle AOP=\angle MOD$           (Vertically opposite angles)

$\triangle APO\, \sim \, \triangle DMO,$       (AA similarity)

$\frac{AP}{DM}=\frac{AO}{DO}$

Since

$\frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{\frac{1}{2}\times BC\times AP}{\frac{1}{2}\times BC\times MD}$

$\Rightarrow \frac{ar(\triangle ABC)}{ar(\triangle BCD)}=\frac{AP}{ MD}=\frac{AO}{DO}$

Let $\triangle ABC\, \sim \, \triangle DEF,$ , therefore,

$ar(\triangle ABC\,) = \,ar( \triangle DEF)$                (Given )

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}................................1$

$\therefore \frac{ar(\triangle ABC)}{ar(\triangle DEF)}=1$

$\Rightarrow \frac{AB^2}{DE^2}=\frac{BC^2}{EF^2}=\frac{AC^2}{DF^2}=1$

$AB=DE$

$BC=EF$

$AC=DF$

$\triangle ABC\, \cong \, \triangle DEF$                    (SSS )

D, E, and F are respectively the mid-points of sides AB, BC and CA of $\Delta ABC$.     ( Given )

$DE=\frac{1}{2}AC$       and          DE||AC

In $\Delta BED \: \:and \: \: \Delta ABC$,

$\angle BED=\angle BCA$                (corresponding angles )

$\angle BDE=\angle BAC$               (corresponding angles )

$\Delta BED \: \:\sim \: \: \Delta ABC$          (By AA)

$\frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{DE^2}{AC^2}$

$\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{(\frac{1}{2}AC)^2}{AC^2}$

$\Rightarrow \frac{ar(\triangle BED)}{ar(\triangle ABC)}=\frac{1}{4}$

$\Rightarrow ar(\triangle BED)=\frac{1}{4}\times ar(\triangle ABC)$

Let ${ar(\triangle ABC)$  be x.

$\Rightarrow ar(\triangle BED)=\frac{1}{4}\times x$

Similarly,

$\Rightarrow ar(\triangle CEF)=\frac{1}{4}\times x$                       and          $\Rightarrow ar(\triangle ADF)=\frac{1}{4}\times x$

$ar(\triangle ABC)=ar(\triangle ADF)+ar(\triangle BED)+ar(\triangle CEF)+ar(\triangle DEF)$

$\Rightarrow x=\frac{x}{4}+\frac{x}{4}+\frac{x}{4}+ar(\triangle DEF)$

$\Rightarrow x=\frac{3x}{4}+ar(\triangle DEF)$

$\Rightarrow x-\frac{3x}{4}=ar(\triangle DEF)$

$\Rightarrow \frac{4x-3x}{4}=ar(\triangle DEF)$

$\Rightarrow \frac{x}{4}=ar(\triangle DEF)$

$\frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{\frac{x}{4}}{x}$

$\Rightarrow \frac{ar(\triangle DEF)}{ar(\triangle ABC)}=\frac{1}{4}$

Let AD and PS be medians of both similar triangles.

$\triangle ABC\sim \triangle PQR$

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}............................1$

$\angle A=\angle P,\angle B=\angle Q,\angle \angle C=\angle R..................2$

$BD=CD=\frac{1}{2}BC\, \, and\, QS=SR=\frac{1}{2}QR$

Purring these value in 1,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AC}{PR}..........................3$

In $\triangle ABD\, and\, \triangle PQS,$

$\angle B=\angle Q$    (proved above)

$\frac{AB}{PQ}=\frac{BD}{QS}$      (proved above)

$\triangle ABD\, \sim \triangle PQS$     (SAS )

Therefore,

$\frac{AB}{PQ}=\frac{BD}{QS}=\frac{AD}{PS}................4$

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AB^2}{PQ^2}=\frac{BC^2}{QR^2}=\frac{AC^2}{PR^2}$

From 1 and 4, we get

$\frac{AB}{PQ}=\frac{BC}{QR}=\frac{AC}{PR}=\frac{AD}{PS}$

$\frac{ar(\triangle ABC)}{ar(\triangle PQR)}=\frac{AD^2}{PS^2}$

Let ABCD be a square of side units.

Therefore, diagonal = $\sqrt{2}a$

Triangles form on the side and diagonal are $\triangle$ ABE and $\triangle$DEF, respectively.

Length of each side of triangle ABE = a units

Length of each side of triangle DEF = $\sqrt{2}a$ units

Both the triangles are equilateral triangles with each angle of $60 \degree$.

$\triangle ABE\sim \triangle DBF$      ( By AAA)

Using area theorem,

$\frac{ar(\triangle ABC)}{ar(\triangle DBF)}=(\frac{a}{\sqrt{2}a})^2=\frac{1}{2}$

(A) 2: 1        (B) 1: 2        (C) 4 : 1        (D) 1: 4

Given: ABC and BDE are two equilateral triangles such that D is the mid-point of BC.

All angles of the triangle are $60 \degree$.

$\triangle$ABC $\sim \triangle$ BDE    (By AAA)

Let AB=BC=CA = x

then   EB=BD=ED=$\frac{x}{2}$

$\frac{ar(\triangle ABC)}{ar(\triangle BDE)}=(\frac{x}{\frac{x}{2}})^2=\frac{4}{1}$

Option C is correct.

(A) 2 : 3        (B) 4: 9        (C) 81: 16        (D) 16: 81

Sides of two similar triangles are in the ratio 4: 9.

Let triangles be ABC and DEF.

We know that

$\frac{ar(\triangle ABC)}{ar(\triangle DEF)}=\frac{AB^2}{DE^2}=\frac{4^2}{9^2}=\frac{16}{81}$

Option D is correct.

NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6.5

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 7 cm, 24 cm

By Pythagoras theorem,

$h^2=7^2+24^2$

$h^2=49+576$

$h^2=625$

$h=25$= given third side.

Hence, it is the right triangle with h=25 cm.

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 3 cm, 6 cm

By Pythagoras theorem,

$h^2=3^2+6^2$

$h^2=9+36$

$h^2=45$

$h=\sqrt{45}\neq 8$

Hence, it is not the right triangle.

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 50 cm, 80 cm

By Pythagoras theorem,

$h^2=50^2+80^2$

$h^2=2500+6400$

$h^2=8900$

$h=\sqrt{8900}\neq 100$

Hence, it is not a right triangle.

In the case of a right triangle, the length of its hypotenuse is highest.

hypotenuse be h.

Taking, 5cm, 12 cm

By Pythagoras theorem,

$h^2=5^2+12^2$

$h^2=25+144$

$h^2=169$

$h=13$= given third side.

Hence, it is a right triangle with h=13 cm.

Let $\angle MPR$   be x

In $\triangle MPR$,

$\angle MRP=180 \degree-90 \degree-x$

$\angle MRP=90 \degree-x$

Similarly,

In $\triangle MPQ$,

$\angle MPQ=90 \degree-\angle MPR$

$\angle MPQ=90 \degree-x$

$\angle MQP=180 \degree-90 \degree-(90 \degree-x)=x$

In $\triangle QMP\, and\, \triangle PMR,$

$\angle MPQ\, =\angle MRP$

$\angle PMQ\, =\angle RMP$

$\angle MQP\, =\angle MPR$

$\triangle QMP\, \sim \triangle PMR,$                (By AAA)

$\frac{QM}{PM}=\frac{MP}{MR}$

$\Rightarrow PM^2=MQ\times MR$

Hence proved.

In $\triangle ADB\, and\, \triangle ABC,$

$\angle DAB\, =\angle ACB \, \, \, \, \, \, \, \, (Each 90 \degree)$

$\angle ABD\, =\angle CBA$       (common )

$\triangle ADB\, \sim \triangle ABC$                (By AA)

$\Rightarrow \frac{AB}{BC}=\frac{BD}{AB}$

$\Rightarrow AB^2=BC.BD$,  hence prooved.

Let $\angle CAB$   be x

In $\triangle ABC$,

$\angle CBA=180 \degree-90 \degree-x$

$\angle CBA=90 \degree-x$

Similarly,

In $\triangle CAD$,

$\angle CAD=90 \degree-\angle CAB$

$\angle CAD=90 \degree-x$

$\angle CDA=180 \degree-90 \degree-(90 \degree-x)=x$

In $\triangle ABC\, and\, \triangle ACD,$

$\angle CBA\, =\angle CAD$

$\angle CAB\, =\angle CDA$

$\angle ACB\, =\angle DCA$         ( Each right angle)

$\triangle ABC\, \sim \triangle ,ACD$                (By AAA)

$\frac{AC}{DC}=\frac{BC}{AC}$

$\Rightarrow AC^2=BC\times DC$

Hence proved

In $\triangle ACD\, and\, \triangle ABD,$

$\angle DCA\, =\angle DAB\, \, \, \, \, \, \, \, (Each 90 \degree)$

$\angle CDA\, =\angle ADB$       (common )

$\triangle ACD\, \sim \triangle ABD$                (By AA)

$\Rightarrow \frac{CD}{AD}=\frac{AD}{BD}$

$\Rightarrow AD^2=BD\times CD$

Hence proved.

Given: ABC is an isosceles triangle right angled at C.

Let AC=BC

In $\triangle$ ABC,

By Pythagoras theorem

$AB^2=AC^2+BC^2$

$AB^2=AC^2+AC^2$           (AC=BC)

$AB^2=2.AC^2$

Hence proved.

Given: ABC is an isosceles triangle with  AC=BC.

In $\triangle$ ABC,

$AB^2=2.AC^2$                       (Given )

$AB^2=AC^2+AC^2$           (AC=BC)

$AB^2=AC^2+BC^2$

These sides satisfy Pythagoras theorem so ABC is a right-angled triangle.

Hence proved.

Given: ABC is an equilateral triangle of side 2a.

AB=BC=AC=2a

We know that the altitude of an equilateral triangle bisects the opposite side.

So, BD=CD=a

In $\triangle$ ADB,

By Pythagoras theorem,

$AB^2=AD^2+BD^2$

$\Rightarrow (2a)^2=AD^2+a^2$

$\Rightarrow 4a^2=AD^2+a^2$

$\Rightarrow 4a^2-a^2=AD^2$

$\Rightarrow 3a^2=AD^2$

$\Rightarrow AD=\sqrt{3}a$

The length of each altitude is $\sqrt{3}a$.

In $\triangle$ AOB, by Pythagoras theorem,

$AB^2=AO^2+BO^2..................1$

In $\triangle$ BOC, by Pythagoras theorem,

$BC^2=BO^2+CO^2..................2$

In $\triangle$ COD, by Pythagoras theorem,

$CD^2=CO^2+DO^2..................3$

In $\triangle$ AOD, by Pythagoras theorem,

$AD^2=AO^2+DO^2..................4$

$AB^2+BC^2+CD^2+AD^2=AO^2+BO^2+BO^2+CO^2+CO^2+DO^2+AO^2+DO^2$

$AB^2+BC^2+CD^2+AD^2=2(AO^2+BO^2+CO^2+DO^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=2(2.AO^2+2.BO^2)$        (AO=CO  and BO=DO)

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4(AO^2+BO^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC}{2})^2+(\frac{BD}{2})^2)$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=4((\frac{AC^2}{4})+(\frac{BD^2}{4}))$

$\Rightarrow AB^2+BC^2+CD^2+AD^2=AC^2+BD^2$

Hence proved.

Join AO, BO, CO

In $\triangle$ AOF, by Pythagoras theorem,

$OA^2=OF^2+AF^2..................1$

In $\triangle$ BOD, by Pythagoras theorem,

$OB^2=OD^2+BD^2..................2$

In $\triangle$ COE, by Pythagoras theorem,

$OC^2=OE^2+EC^2..................3$

$OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2$$\Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4$

Hence proved

Join AO, BO, CO

In $\triangle$ AOF, by Pythagoras theorem,

$OA^2=OF^2+AF^2..................1$

In $\triangle$ BOD, by Pythagoras theorem,

$OB^2=OD^2+BD^2..................2$

In $\triangle$ COE, by Pythagoras theorem,

$OC^2=OE^2+EC^2..................3$

$OA^2+OB^2+OC^2=OF^2+AF^2+OD^2+BD^2+OE^2+EC^2$$\Rightarrow OA^2+OB^2+OC^2-OD^2-OE^2-OF^2=AF^2+BD^2+EC^2....................4$

$\Rightarrow (OA^2-OE^2)+(OC^2-OD^2)+(OB^2-OF^2)=AF^2+BD^2+EC^2$$\Rightarrow AE^2+CD^2+BF^2=AF^2+BD^2+EC^2$

OA is a wall and AB is a ladder.

In $\triangle$ AOB, by Pythagoras theorem

$AB^2=AO^2+BO^2$

$\Rightarrow 10^2=8^2+BO^2$

$\Rightarrow 100=64+BO^2$

$\Rightarrow 100-64=BO^2$

$\Rightarrow 36=BO^2$

$\Rightarrow BO=6 m$

Hence, the distance of the foot of the ladder from the base of the wall is 6 m.

OB is a pole.

In $\triangle$ AOB, by Pythagoras theorem

$AB^2=AO^2+BO^2$

$\Rightarrow 24^2=18^2+AO^2$

$\Rightarrow 576=324+AO^2$

$\Rightarrow 576-324=AO^2$

$\Rightarrow 252=AO^2$

$\Rightarrow AO=6\sqrt{7} m$

Hence, the distance of the stack from the base of the pole is $6\sqrt{7}$ m.

Distance travelled by the first aeroplane due north in $1\frac{1}{2}$ hours.

$=1000\times \frac{3}{2}=1500 km$

Distance travelled by second aeroplane due west in $1\frac{1}{2}$ hours.

$=1200\times \frac{3}{2}=1800 km$

OA and OB are the distance travelled.

By Pythagoras theorem,

$AB^2=OA^2+OB^2$

$\Rightarrow AB^2=1500^2+1800^2$

$\Rightarrow AB^2=2250000+3240000$

$\Rightarrow AB^2=5490000$

$\Rightarrow AB^2=300\sqrt{61}km$

Thus, the distance between the two planes is $300\sqrt{61}km$.

Let  AB and CD be poles of heights 6 m and 11 m respectively.

CP=11-6=5 m     and    AP= 12 m

In $\triangle$ APC,

By Pythagoras theorem,

$AP^2+PC^2=AC^2$

$\Rightarrow 12^2+5^2=AC^2$

$\Rightarrow 144+25=AC^2$

$\Rightarrow 169=AC^2$

$\Rightarrow AC=13m$

Hence, the distance between the tops of two poles is 13 m.

In $\triangle$ ACE, by Pythagoras theorem,

$AE^2=AC^2+CE^2..................1$

In $\triangle$ BCD, by Pythagoras theorem,

$DB^2=BC^2+CD^2..................2$

From 1 and 2, we get

$AC^2+CE^2+BC^2+CD^2=AE^2+DB^2..................3$

In $\triangle$ CDE, by Pythagoras theorem,

$DE^2=CD^2+CE^2..................4$

In $\triangle$ ABC, by Pythagoras theorem,

$AB^2=AC^2+CB^2..................5$

From 3,4,5 we get

$DE^2+AB^2=AE^2+DB^2$

In $\triangle$ ACD, by Pythagoras theorem,

$AC^2=AD^2+DC^2$

$AC^2-DC^2=AD^2..................1$

In $\triangle$ ABD, by Pythagoras theorem,

$AB^2=AD^2+BD^2$

$AB^2-BD^2=AD^2.................2$

From 1 and 2, we get

$AC^2-CD^2=AB^2-DB^2..................3$

Given : 3DC=DB, so

$CD=\frac{BC}{4}\, \, and\, \, BD=\frac{3BC}{4}........................4$

From 3 and 4, we get

$AC^2-(\frac{BC}{4})^2=AB^2-(\frac{3BC}{4})^2$

$AC^2-(\frac{BC^2}{16})=AB^2-(\frac{9BC^2}{16})$

$16AC^2-BC^2=16AB^2- 9BC^2$

$16AC^2=16AB^2- 8BC^2$

$\Rightarrow 2AC^2=2AB^2- BC^2$

$2 AB^2 = 2 AC^2 + BC^2.$

Hence proved.

Given: An equilateral triangle ABC, D is a point on side BC such that BD = 1/3  BC.

To prove :  $9 AD^2 = 7 AB^2$

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, $BE=CE=\frac{a}{2}$

In $\triangle$AEB, by Pythagoras theorem

$AB^2=AE^2+BE^2$

$a^2=AE^2+(\frac{a}{2})^2$

$\Rightarrow a^2-(\frac{a^2}{4})=AE^2$

$\Rightarrow (\frac{3a^2}{4})=AE^2$

$\Rightarrow AE=(\frac{\sqrt{3}a}{2})$

Given :  BD = 1/3  BC.

$BD=\frac{a}{3}$

$DE=BE=BD=\frac{a}{2}-\frac{a}{3}=\frac{a}{6}$

In $\triangle$ADE, by Pythagoras theorem,

$AD^2=AE^2+DE^2$

$\Rightarrow AD^2=(\frac{\sqrt{3}a}{2})^2+(\frac{a}{6})^2$

$\Rightarrow AD^2=(\frac{3a^2}{4})+(\frac{a^2}{36})$

$\Rightarrow AD^2=(\frac{7a^2}{9})$

$\Rightarrow AD^2=(\frac{7AB^2}{9})$

$\Rightarrow 9AD^2=7AB^2$

Given: An equilateral triangle ABC.

Let AB=BC=CA=a

Draw an altitude AE on BC.

So, $BE=CE=\frac{a}{2}$

In $\triangle$AEB, by Pythagoras theorem

$AB^2=AE^2+BE^2$

$a^2=AE^2+(\frac{a}{2})^2$

$\Rightarrow a^2-(\frac{a^2}{4})=AE^2$

$\Rightarrow (\frac{3a^2}{4})=AE^2$

$\Rightarrow 3a^2=4AE^2$

$\Rightarrow 4.(altitude)^2=3.(side)^2$

(B) 60°

(C) 90°

(D) 45°

In $\Delta ABC$  AB = $6 \sqrt 3$  cm, AC = 12 cm and BC = 6 cm.

$AB^2+BC^2=108+36$

$=144$

$=12^2$

$=AC^2$

It satisfies the Pythagoras theorem.

Hence, ABC is a right-angled triangle and right-angled at B.

Option C is correct.

NCERT solutions for class 10 maths chapter 6 Triangles Excercise: 6.6

A line RT is drawn parallel to SP which intersect QP produced at T.

Given: PS is the bisector of $\angle QPR \: \: of\: \: \Delta PQR$.

$\angle QPS=\angle SPR.....................................1$

By construction,

$\angle SPR=\angle PRT.....................................2$   (as PS||TR)

$\angle QPS=\angle QTR.....................................3$   (as PS||TR)

From the above equations, we get

$\angle PRT=\angle QTR$

$\therefore PT=PR$

By construction, PS||TR

In $\triangle$QTR, by Thales theorem,

$\frac{QS}{SR}=\frac{QP}{PT}$

$\frac{QS }{SR } = \frac{PQ }{PR }$

Hence proved.

Join BD

Given :  D is a point on hypotenuse AC of D ABC, such that BD $\perp$ AC, DM $\perp$ BC and  DN $\perp$ AB.Also DN || BC, DM||NB

$\angle CDB=90 \degree$

$\Rightarrow \angle 2+\angle 3=90 \degree.............................1$

In $\triangle$ CDM, $\angle 1+\angle 2+\angle DMC=180 \degree$

$\angle 1+\angle 2=90 \degree.......................2$

In $\triangle$ DMB, $\angle 3+\angle 4+\angle DMB=180 \degree$

$\angle 3+\angle 4=90 \degree.......................3$

From equation 1 and 2, we get $\angle 1=\angle 3$

From equation 1 and 3, we get $\angle 2=\angle 4$

In $\triangle DCM\, \, and\, \, \triangle BDM,$

$\angle 1=\angle 3$

$\angle 2=\angle 4$

$\triangle DCM\, \, \sim \, \, \triangle BDM,$     (By AA)

$\Rightarrow \frac{BM}{DM}=\frac{DM}{MC}$

$\Rightarrow \frac{DN}{DM}=\frac{DM}{MC}$           (BM=DN)

$\Rightarrow$$DM^2 = DN . MC$

Hence proved

In $\triangle$ DBN,

$\angle 5+\angle 7=90 \degree.......................1$

In $\triangle$ DAN,

$\angle 6+\angle 8=90 \degree.......................2$

BD $\perp$AC, $\therefore \angle ADB=90 \degree$

$\angle 5+\angle 6=90 \degree.......................3$

From equation 1 and 3, we get $\angle 6=\angle 7$

From equation 2 and 3, we get $\angle 5=\angle 8$

In $\triangle DNA\, \, and\, \, \triangle BND,$

$\angle 6=\angle 7$

$\angle 5=\angle 8$

$\triangle DNA\, \, \sim \, \, \triangle BND$     (By AA)

$\Rightarrow \frac{AN}{DN}=\frac{DN}{NB}$

$\Rightarrow \frac{AN}{DN}=\frac{DN}{DM}$          (NB=DM)

$\Rightarrow$$DN^2 = AN . DM$

Hence proved.

In $\triangle$ADB, by Pythagoras theorem

$AB^2=AD^2+DB^2.......................1$

In $\triangle$ACD, by Pythagoras theorem

$AC^2=AD^2+DC^2.......................2$

$AC^2=AD^2+(BD+BC)^2$

$\Rightarrow AC^2=AD^2+(BD)^2+(BC)^2+2.BD.BC$

$AC^2 = AB^2 + BC^2 + 2 BC . BD.$            (From 1)

In $\triangle$ADB, by Pythagoras theorem

$AB^2=AD^2+DB^2$

$AD^2=AB^2-DB^2...........................1$

In $\triangle$ACD, by Pythagoras theorem

$AC^2=AD^2+DC^2$

$AC^2=AB^2-BD^2+DC^2$         (From 1)

$\Rightarrow AC^2=AB^2-BD^2+(BC-BD)^2$

$\Rightarrow AC^2=AB^2-BD^2+(BC)^2+(BD)^2-2.BD.BC$

$AC^2 = AB^2 + BC^2 - 2 BC . BD.$

Given: AD is a median of a triangle ABC and  AM $\perp$ BC.

In $\triangle$AMD, by Pythagoras theorem

$AD^2=AM^2+MD^2.......................1$

In $\triangle$AMC, by Pythagoras theorem

$AC^2=AM^2+MC^2$

$AC^2=AM^2+(MD+DC)^2$

$\Rightarrow AC^2=AM^2+(MD)^2+(DC)^2+2.MD.DC$

$AC^2 = AD^2 + DC^2 + 2 DC . MD.$            (From 1)

$AC^2 = AD^2 + (\frac{BC}{2})^2 + 2(\frac{BC}{2}). MD.$           (BC=2 DC)

$AC ^2 = AD ^2 + BC DM + \left ( \frac{BC}{2} \right ) ^2$

In $\triangle$ABM, by Pythagoras theorem

$AB^2=AM^2+MB^2$

$AB^2=(AD^2-DM^2)+MB^2$

$\Rightarrow AB^2=(AD^2-DM^2)+(BD-MD)^2$

$\Rightarrow AB^2=AD^2-DM^2+(BD)^2+(MD)^2-2.BD.MD$

$\Rightarrow AB^2=AD^2+(BD)^2-2.BD.MD$

$\Rightarrow AB^2 = AD^2 + (\frac{BC}{2})^2 -2(\frac{BC}{2}). MD.=AC^2$           (BC=2 BD)

$\Rightarrow AD^2 + (\frac{BC}{2})^2 -BC. MD.=AC^2$

In $\triangle$ABM, by Pythagoras theorem

$AB^2=AM^2+MB^2.......................1$

In $\triangle$AMC, by Pythagoras theorem

$AC^2=AM^2+MC^2$..................................2

$AB^2+AC^2=2AM^2+MB^2+MC^2$

$\Rightarrow AB^2+AC^2=2AM^2+(BD-DM)^2+(MD+DC)^2$

$\Rightarrow AB^2+AC^2=2AM^2+(BD)^2+(DM)^2-2.BD.DM+(MD)^2+(DC)^2+2.MD.DC$

$\Rightarrow AB^2+AC^2=2AM^2+2.(DM)^2+BD^2+(DC)^2+2.MD.(DC-BD)$$\Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2+2.MD.(\frac{BC}{2}-\frac{BC}{2})$$\Rightarrow AB^2+AC^2=2(AM^2+(DM)^2)+(\frac{BC}{2})^2+(\frac{BC}{2})^2$

$AC ^2 + AB ^2 = 2 AD^2 + \frac{1}{2} BC ^2$

In parallelogram ABCD, AF and DE are altitudes drawn on DC and produced BA.

In $\triangle$DEA, by Pythagoras theorem

$DA^2=DE^2+EA^2.......................1$

In $\triangle$DEB, by Pythagoras theorem

$DB^2=DE^2+EB^2$

$DB^2=DE^2+(EA+AB)^2$

$DB^2=DE^2+(EA)^2+(AB)^2+2.EA.AB$

$DB^2=DA^2+(AB)^2+2.EA.AB$....................................2

In $\triangle$ADF, by Pythagoras theorem

$DA^2=AF^2+FD^2$

In $\triangle$AFC, by Pythagoras theorem

$AC^2=AF^2+FC^2=AF^2+(DC-FD)^2$

$\Rightarrow AC^2=AF^2+(DC)^2+(FD)^2-2.DC.FD$

$\Rightarrow AC^2=(AF^2+FD^2)+(DC)^2-2.DC.FD$

$\Rightarrow AC^2=AD^2+(DC)^2-2.DC.FD.......................3$

Since ABCD is a parallelogram.

In $\triangle DEA\, and\, \triangle ADF,$

$\angle DEA=\angle AFD\, \, \, \, \, \, \, (each 90 \degree)$

$\angle DAE=\angle ADF$      (AE||DF)

$\triangle DEA\, \cong \, \triangle ADF,$        (ASA rule)

$\Rightarrow EA=DF.......................6$

$DA^2+AB^2+2.EA.AB+AD^2+DC^2-2.DC.FD=DB^2+AC^2$$\Rightarrow DA^2+AB^2+AD^2+DC^2+2.EA.AB-2.DC.FD=DB^2+AC^2$

$\Rightarrow BC^2+AB^2+AD^2+2.EA.AB-2.AB.EA=DB^2+AC^2$   (From 4 and 6)

$\Rightarrow BC^2+AB^2+CD^2=DB^2+AC^2$\

Join BC

In $\triangle APC\, \, and\, \triangle DPB,$

$\angle APC\, \, = \angle DPB$        ( vertically opposite angle)

$\angle CAP\, \, = \angle BDP$      (Angles in the same segment)

$\triangle APC\, \, \sim \triangle DPB$         (By AA)

Join BC

In $\triangle APC\, \, and\, \triangle DPB,$

$\angle APC\, \, = \angle DPB$        ( vertically opposite angle)

$\angle CAP\, \, = \angle BDP$      (Angles in the same segment)

$\triangle APC\, \, \sim \triangle DPB$         (By AA)

$\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$             (Corresponding sides of similar triangles are proportional)

$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$

$\Rightarrow AP.PB=PC.DP$

In $\Delta PAC \,and \,\Delta PDB,$

$\angle P=\angle P$         (Common)

$\angle PAC=\angle PDB$  (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So,  $\Delta PAC \sim \Delta PDB$       ( By AA rule)

In $\Delta PAC \,and \,\Delta PDB,$

$\angle P=\angle P$         (Common)

$\angle PAC=\angle PDB$  (Exterior angle of a cyclic quadrilateral is equal to opposite interior angle)

So,  $\Delta PAC \sim \Delta PDB$       ( By AA rule)

24440$\frac{AP}{DP}=\frac{PC}{PB}=\frac{CA}{BD}$             (Corresponding sides of similar triangles are proportional)

$\Rightarrow \frac{AP}{DP}=\frac{PC}{PB}$

$\Rightarrow AP.PB=PC.DP$

$\frac{BD }{CD} = \frac{AB}{AC}$         (Given )
$\Rightarrow \frac{BD }{CD} = \frac{AP}{AC}$