Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
Filter By
Clear Filter

All Questions

P is the mid-point of the side CD of a parallelogram ABCD. A line through C parallel to PA intersects AB at Q and DA produced at R. Prove that DA = AR and CQ = QR.

Solution.
Given: In a parallelogram ABCD, P is the mid-point of DC
To Prove: DA = AR and CQ = QR

Proof: ABCD is a parallelogram
 BC = AD and BC\parallel AD
Also, DC = AB and DC\parallel AB
P is mid-point of DC
DP=PC=\frac{1}{2}DC
Now QC\parallel AP and PC\parallel AQ
So APCQ is a parallelogram
AQ=PC=\frac{1}{2}DC

\frac{1}{2}AB=BQ         …..(1) {\because DC = AB}
In \triangle AQR  &  \triangle BQC AQ = BQ {from equation 1}
\angle AQR=\angle BQR {vertically opposite angles}
\angle ARQ=\angle BCQ {alternate angles of transversal}
\triangle AQR\cong \triangle BQC {AAS congruence}
AR = BC {by CPCT}
BC = DA
AR = DA
Also, CQ = QR {by CPCT}
Hence Proved .

View Full Answer(1)
Posted by

infoexpert26

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium.

Solution.

Given: Let ABCD be a trapezium in which AB\parallel DC  and let M and N be the mid-points of diagonals AC and BD.

To Prove: MN\parallel AB\parallel CD
Proof: Join CN and produce it to meet AB at E
In \triangle CDN and \triangle EBN we have
DN = BN {N is mid-point of BD}
\angle DCN=\angle BEN {alternate interior angle}
\angle CDN=\angle EBN {alternate interior angles}
\triangle CDN\cong \triangle EBN 
DC = EB and CN = NE {by CPCT}
Thus in \triangle CAE, the points M and N are the mid-points of AC and CE, respectively.

MN\parallel AE   {By mid-point theorem}
MN\parallel AB\parallel CD
Hence Proved

View Full Answer(1)
Posted by

infoexpert26

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

D, E and F are respectively the mid-points of the sides AB, BC and CA of a triangle ABC. Prove that by joining these mid-points D, E and F, the triangles ABC is divided into four congruent triangles.

Solution.
 

Given: In \triangle ABC, D, E and F are respectively the mid-points of the sides AB, BC and CA.
To prove: \triangle ABC is divided into four congruent triangles.

Proof: Using given conditions we have
AD=BD=\frac{1}{2}AB,BE=EC=\frac{1}{2}BC,AF=CF=\frac{1}{2}AC
Using mid-point theorem
EF\parallel AB  and   EF=\frac{1}{2}AB=AD=BD
ED\parallel AC  and  ED=\frac{1}{2}AC=AF=CF
DF\parallel BC  and   DF=\frac{1}{2}BC=BE=EC
In \triangle ADF and \triangle EFD
 AD = EF
 AF = DE
 DF = FD {common side}
\triangle ADF\cong \triangle EFD  {by SSS congruence}
Similarly we can prove that,
\triangle DEF\cong \triangle EDB
\triangle DEF\cong \triangle CFE
So, \triangle ABC is divided into four congruent triangles
Hence proved

View Full Answer(1)
Posted by

infoexpert26

ABCD is a rectangle in which diagonal BD bisects \angle B. Show that ABCD is a square. 

Solution.
Given: In a rectangle ABCD, diagonal BD bisects B
To Prove: ABCD is a square
Construction: Join AC

Proof:
Given that ABCD is a rectangle. So all angles are equal to 90^{\circ}
Now, BD bisects \angle B
\angle DBA=\angle CBD
Also,
\angle DBA+\angle CBD=90^{\circ}  
So,
2\angle DBA=90^{\circ}


\angle DBA=45^{\circ}
In \triangle ABD,
\angle ABD+\angle BDA+\angle DAB=180^{\circ}  
(Angle sum property)
45^{\circ}+\angle BDA+90^{\circ}=180^{\circ}
\angle BDA=45^{\circ}
In \triangle ABD,
AD = AB (sides opposite to equal angles in a triangle are equal)
Similarly, we can prove that BC = CD
So, AB = BC = CD = DA
So ABCD is a square.
Hence proved

View Full Answer(1)
Posted by

infoexpert26

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

P and Q are points on opposite sides AD and BC of a parallelogram ABCD such that PQ passes through the point of intersection O of its diagonals AC and BD. Show that PQ is bisected at O.

Solution.

Given: ABCD is a parallelogram whose diagonals bisect each other at O.
To Prove: PQ is bisected at O.


Proof: In \triangle ODP and \triangle OBQ
\angle BOQ=\angle POD {Vertically opposite angles}
\angle OBQ=\angle ODP {interior angles}
OB = OD {given}
 \triangle ODP\cong \triangle OBQ {by ASA congruence}
OP = OQ {by CPCT rule}
So, PQ is bisected at O

 

View Full Answer(1)
Posted by

infoexpert26

Prove that the quadrilateral formed by the bisectors of the angles of a parallelogram is a rectangle.

Solution.
Given: Let ABCD be a parallelogram and AP, BR, CQ, DS are the bisectors of
\angle A, \angle B, \angle Cand \angle D respectively.
To Prove: Quadrilateral PQRS is a rectangle.

Proof: Since ABCD is a parallelogram
Then DC\parallel AB and DA is transversal.
\angle A+\angle B=180^{\circ}{sum of co-interior angles of a parallelogram}
\frac{1}{2}\angle A+\frac{1}{2}\angle B=90^{\circ} {Dividing both sides by 2}
\angle PAD+\angle PDA=90^{\circ}
\Rightarrow \angle APD=90^{\circ} {\because sum of all the angles of a triangle is 1800}
 \angle QRS=90^{\circ}
Similarly,
\angle RBC+\angle RCB=90^{\circ}
\Rightarrow \angle BRC=90^{\circ}  
\angle QRS=90^{\circ}
Similarly,
\angle QAB+\angle QBA=90^{\circ}
\Rightarrow \angle AQB=90^{\circ}  {\because sum of all the angles of a triangle is 1800}
\angle RQP=90^{\circ} {\because vertically opposite angles}
Similarly,|

\angle SDC+\angle SCD=90^{\circ}
\Rightarrow \angle DSC=90^{\circ} {\because sum of all the angles of a triangle is 1800}
 \angle RSP=90^{\circ} {\because vertically opposite angles}
Thus PQRS is a quadrilateral whose all angles are 90^{\circ}
Hence PQRS is a rectangle.
Hence proved

View Full Answer(1)
Posted by

infoexpert26

NEET 2024 Most scoring concepts

    Just Study 32% of the NEET syllabus and Score up to 100% marks

E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD. Prove that EF \parallel AB and EF EF=\frac{1}{2}(AB+CD)

Solution.
Given: ABCD is a trapezium in which AB\parallel CD, E and F are the mid-points or sides AD and BC.


Constructions: Joint BE and produce it to meet CD at G.
Draw BOD which intersects EF at O
To Prove: EF \parallel AB  and EF=\frac{1}{2}(AB+CD)

Proof: In \triangle GCB, E and F are respectively the mid-points of BG and BC, then by mid-point theorem.
EF \parallel GC
But, GC\parallel AB or  CD \parallel AB           {given}
\therefore EF\parallel AB
In \triangle ADB, AB \parallel EO and E is the mid-point of AD. Then by mid-point theorem, O is mid-point of BD.
EO=\frac{1}{2}AB          ……(1)
In \triangle BDC, OF \parallel CDand O is the mid-point of BD
OF=\frac{1}{2}CD         …..(2)
Adding 1 and 2, we get
EO+OF=\frac{1}{2}AB+\frac{1}{2}CD
EF=\frac{1}{2}(AB+CD)

Hence Proved

View Full Answer(1)
Posted by

infoexpert26

Show that the quadrilateral formed by joining the mid-points of the consecutive sides of a square is also a square.
Hint:  Prove all the sides are equal and diagonals are equal 

Solution.
Given: In a square ABCD; P, Q, R and S are the mid-points of AB, BC, CD and DA.
To Prove: PQRS is a square
Construction: Join AC and BD

Proof : Here ABCD is a square
\ AB = BC = CD = AD
P, Q, R, S are the mid-points of AB, BC, CD and DA.
In \triangle ADC,
SR\parallel AC
SR=\frac{1}{2}AC  {using mid-point theorem} …..(1)
In \triangle ABC,
PQ\parallel AC
PQ=\frac{1}{2}AC {using mid-point theorem} …..(2)
From equation 1 and 2
SR\parallel PQ  and SR= PQ=\frac{1}{2}AC    …..(3)
Similarly, SP\parallel BD  and BD\parallel RQ
\ SP=\frac{1}{2}BD and  RQ=\frac{1}{2}BD   {using mid-point theorem}
 SP=RQ=\frac{1}{2}BD
Since diagonals of a square bisect each other at right angles.
 AC = BD
\Rightarrow SP=RQ=\frac{1}{2}AC …..(4)
From equation 3 and 4
SP = PQ = SP = RQ {All sides are equal}
In quadrilateral OERF
OE\parallel FR and OF\parallel ER
\angle EOF=\angle ERF=90^{\circ}
In quadrilateral RSPQ
\angle SRQ=\angle ERF=\angle SPQ=90^{\circ}   {Opposite angles in a parallelogram are equal}
\angle RSP + \angle SRQ + \angle RQP + \angle QRS = 360^{\circ}
90^{\circ} +90^{\circ}+ \angle RSP + \angle RQP = 360^{\circ}
\angle RSP + \angle RQP = 180^{\circ}
\angle RSP = \angle RQP = 90^{\circ}
Since all the sides are equal and angles are also equal, so PQRS is a square.
Hence Proved

View Full Answer(1)
Posted by

infoexpert26

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

E is the mid-point of a median AD of \triangle ABC and BE is produced to meet AC at F. Show that AF=\frac{1}{3}AC .

Solution.
Given: In \triangle ABC, AD is a median and E is the mid-point of AD
To Prove: AF=\frac{1}{3}AC.
Construction: Draw DP\parallel EF and \triangle ABC as given

Proof: In \triangle ADP, E is mid-point of AD and EF\parallel DB
So, F is mid-point of AP {converse of midpoint theorem}
In \triangle FBC, D is mid-point of BC and DP\parallel BF
So, P is mid-point of FC {converse of midpoint theorem}
Thus, AF = FP = PC
AF+FP+PC=AC=3AF
\therefore AF=\frac{1}{3}AC
Hence Proved

View Full Answer(1)
Posted by

infoexpert26

In Figure, AB \parallel DE, AB = DE, AC \parallel DF and AC = DF. Prove that BC \parallel EF and  BC = EF.

Solution.
Given: AB \parallel DE  and  AC \parallel DF
Also, AB = DE and AC = DF
To prove:  BC\parallel EFand BC = EF

Proof: AB \parallel DE  and AB = DE
AC \parallel DF and AC = DF
In ACFD quadrilateral
AC\parallel FD and AC = FD   …..(1)
Thus ACFD is a parallelogram
AD\parallel CF and AD = CF …..(2)
In ABED quadrilateral
AB\parallel DE and AB = DE   …..(3)
Thus ABED is a parallelogram
AD\parallel BE  and AD = BE …..(4)
From equation 2 and 4
AD = BE = CF and AD \parallel CF \parallel BE …..(5)
In quadrilateral BCFE, BE = CF and BE\parallel CF   {from equation 5}
So, BCFE is a parallelogram BC = EF and BC\parallel EF
Hence Proved.

View Full Answer(1)
Posted by

infoexpert26

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE
filter_img