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#6.4

In fig. , l || m and line segments AB, CD and EF are concurrent at point P. Prove tha

t $\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$ .

Given: l || m and line segments AB, CD and EF are concurrent at point P

To prove:-

$\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$

Proof :In $\Delta APC \; and\; \Delta DPB$

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

$\angle PAC=\angle PBD$ (alternate angles)

and we know that if two angles of one triangle are equal to the two angles of another triangle then the two triangles are similar by AA similarity criterion

$\therefore \; \; \; \; \; \Delta APC\sim \Delta DPB$

Then, $\frac{AP}{BP}=\frac{AC}{BD}=\frac{PC}{PD}\; \; \; \; \; \; \; \; \; ....(1)$

In $\Delta APE\; and\; \Delta FPB$

$\angle APE=\angle BPF$ (vertically opposite angle)

$\angle PAE=\angle PBF$ (alternate angle)

$\therefore \; \; \; \; \Delta APE\sim \Delta FPB$ (by AA similarity criterion)

Then $\frac{AP}{PB}=\frac{AE}{BF}=\frac{PE}{PF}\; \; \; \; \; \; \; ...(2)$

In $\Delta PEC\; and\; \Delta PFD$

$\angle EPC=\angle FPD$ (vertically opposite angle)

$\angle PCE=\angle PDF$ (alternate angle)

$\therefore \; \; \; \; \Delta PEC\sim \Delta PFD$ (by AA similarity criterion)

Then $\frac{PE}{PF}=\frac{PC}{PD}=\frac{EC}{FD}\; \; \; \; \; \; \; ...(3)$

From equation (1), (2) and (3) we get

$\frac{AP}{BP}=\frac{AC}{BD}=\frac{AE}{BF}=\frac{PE}{PF}=\frac{EC}{FD}$

$\Rightarrow \frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$

Hence proved.

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#6.4

In Fig., PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.

Answer: $\left [ PQ=8_{cm},QR=12\; cm,RS=16\; cm\right ]$

Given :- PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm

We know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

$\therefore PA \parallel QB \parallel RC \parallel SD$ (parallel lines)

Then according to basic proportionality theorem

$PQ : QR : RS = AB : BC : CD = 6 : 9 : 12$

$\text {Let}\\PQ = 6x\\ QR = 9x\\ RS = 12x$

length of PS = 36 (given)

$\therefore \; \; \; PQ + QR + RS = 36\; \; \; \Rightarrow \; \; \; 6x + 9x + 12x = 36 \; \; \; \; \; \Rightarrow \; \; \; \; \; 27x = 36$

$x=\frac{36}{27}=\frac{4}{3}$

$PQ=6x=6 \times\frac{4}{3}=8\; cm$

$QR=9x=\frac{9\times 4}{3}=12\; cm$

$RS=12x=\frac{12\times 4}{3}=16\; cm$

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#6.4

O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

ABCD is a trapezium and O is the point of intersection of the diagonals AC and BD.

$AB\parallel DC$

Proof :- $\text {In} \Delta ABD \; \text {and }\; \Delta POD$

$\angle D=\angle D$ (Common angle)

$\angle ABD=\angle POD$ (corresponding angles)

$\therefore \Delta ABD\sim \Delta POD$ (by AA similarity criterion)

Then $\frac{OP}{AB}=\frac{PD}{AD}\; \; \; \; \; \; \; \; \; \; ...(1)$

$\text {In} \Delta ABC \; \text {and }\; \Delta OQC$

$\angle C=\angle C$ (Common angle)

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

$\therefore \Delta ABC\sim \Delta OQC$ (by AA similarity criterion)

Then $\frac{OQ}{AB}=\frac{QC}{BC}\; \; \; \; \; \; \; \; \; \; ...(2)$

$\text {In}\Delta ADC$

$OP\parallel DC$

We know that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

$\therefore \frac{AP}{PD}=\frac{OA}{OC}\; \; \; \; \; \; \; \; \; ...(3)$

$\text {Also In}\; \Delta ABC$

$OQ\parallel AB$

$\therefore \frac{BQ}{QC}=\frac{OA}{OC}\; \; \; \; \; \; \; \; \; ...(4)$ (by basic proportionality theorem)

from equation (3) and (4)

$\frac{AP}{PD}=\frac{BQ}{QC}$

Add 1 on both sides we get

$\frac{AP}{PD}+1=\frac{BQ+QC}{QC}$

$\frac{AP}{PD}=\frac{BC}{QC}$

$\Rightarrow \frac{PD}{AD}=\frac{QC}{BC}\; \; \; \; \; \; \; ...(5)$

$\frac{OP}{AB}=\frac{QC}{BC}$ (use equation (1))

$\Rightarrow \frac{OP}{AB}=\frac{OQ}{AB}$ (use equation (2))

$\Rightarrow OP=OQ$

Hence proved

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#6.4

In Fig., line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and $\angle AEF=\angle AFE$ .Prove that $\frac{BD}{CD}=\frac{BF}{CE}$ .

Given: Line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and $\angle AEF=\angle AFE$

To prove :-

$\frac{BD}{CD}=\frac{BF}{CE}$

Construction:- Take point G on AB such that $CG\parallel DF$

Proof:- E is mid-point of CA (given)

$\therefore \; \; \; \; \; \; CE = AE$

In $\Delta ACG, CG\parallel EF$ and E is mid-point of CA

Then according to mid-point theorem

$CE = GF \; \; \; \; \; \; \; ....(1)$

In $\Delta BCG \; and\; \Delta BDF \; \; CG || DF$

According to basic proportionality theorem.

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

$\therefore \frac{BC}{CD}=\frac{BG}{GF}$

$\frac{BC}{CD}=\frac{BF-GF}{GF}$

$\frac{BC}{CD}=\frac{BF}{GF}=-1$

$\frac{BC}{CD}+1=\frac{BF}{GF}$

$\frac{BC}{CD}+1=\frac{BF}{CE}$ (use equation (1))

$\frac{BC+CD}{CD}=\frac{BF}{CE}$

$\frac{BD}{CD}=\frac{BF}{CE}$

Hence proved.

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#6.4

Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

Let PQR is a right angle triangle which is right angle at point Q.

$PQ = b, QR = a$

Three semicircles are drawn on the sides of $\Delta PQR$ having diameters PQ, QR and PR respectively.

Let $x_{1}, x_{2}$ and $x_{3}$ are the areas of semicircles respectively.

To prove:- $x_{3}=x_{1}+x_{2}$

Proof: In $\Delta PQR$ use Pythagoras theorem we get

$PR^{2}=PQ^{2}+QR^{2}$

$PR^{2}=a^{2}+b^{2}$

$PR^{2}=\sqrt{a^{2}+b^{2}}$

Now area of semi-circle drawn on side PR is

$x_{3}=\frac{\pi }{2}\left ( \frac{PR}{2} \right )^{2}$ $\left [ \therefore \text {area of semicircle=}\frac{\pi r^{2}}{2} \right ]$

$=\frac{\pi}{2}\left ( \frac{\sqrt{a^{2}+b^{2}}}{2} \right )^{2}$ $\left ( \therefore PR=\sqrt{a^{2}+b^{2}} \right )$

$=\frac{\pi}{2}\times\frac{\left ( a^{2}+b^{2} \right )}{4}$

$x_{3}=\frac{\pi}{8}\left ( a^{2}+b^{2} \right )$

area of semi-circle drawn a side QR is

$x_{2}=\frac{\pi }{2}\left ( \frac{QR}{2} \right )^{2}$

$=\frac{\pi }{2}\left ( \frac{a}{2} \right )^{2}$

$=\frac{\pi}{2}\times \frac{a^{2}}{4}$

$x_{2}=\frac{\pi}{8}a^{2}\; \; \; \; \; \; \; ....(2)$

area of semicircle drawn on side PQ is

$x_{1}=\frac{\pi }{2}\left ( \frac{PQ}{2} \right )^{2}$

$x_{1}=\frac{\pi }{2}\left ( \frac{b^{2}}{4} \right )\Rightarrow \frac{\pi}{8}b^{2}\; \; \; \; \; \; \; \; ....(3)$

add equation (2) and (3) we get

$x_{2}+x_{1}=\frac{\pi }{8}a^{2}+\frac{\pi}{8}b^{2}$

$x_{2}+x_{1}=\frac{\pi }{8}\left ( a^{2}+b^{2} \right )=x_{3}$

Hence $x_{2}+x_{1}=x_{3}$

Hence proved

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#6.4

Prove that the area of the equilateral triangle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the equilateral triangles drawn on the other two sides of the triangle.

Let PQR is a right angle triangle which is a right angle at point R.

$PR = b, RQ = a$

Three equilateral triangles are drawn on the sides of triangle PQR that is PRS, RTQ and PUQ

Let $x_{1},x_{2}$ and $x_{3}$ are the areas of equilateral triangles respectively

To prove:- $x_{1}+x_{2}=x_{3}$

Using Pythagoras theorem in $\Delta PQR$ we get

$PQ^{2}+RQ^{2}+RP^{2}$

$PQ^{2}=a^{2}+b^{2}$

$PQ=\sqrt{a^{2}+b^{2}}$

The formula of area of an equilateral triangle is

$=\frac{\sqrt{3}}{4}\left ( side \right )^{2}$

$\therefore$ Area of equilateral $\Delta RTQ$

$x_{1}=\frac{\sqrt{3}}{4}\left ( a^{2} \right )\; \; \; \; \; \; \; \; ....(1)$

Area of equilateral $\Delta RSP$

$x_{2}=\frac{\sqrt{3}}{4} b^{2} \; \; \; \; \; \; \; \; ....(2)$

Area of equilateral $\Delta PQU$

$x_{3}=\frac{\sqrt{3}}{4}\left ( \sqrt{a^{2}+b^{2}} \right )^{2}$ $\left ( Q\; PQ=\sqrt{a^{2}+b^{2}} \right )$

$x_{3}=\frac{\sqrt{3}}{4} \left ( a^{2}+b^{2} \right )$

add equation (1) and (2) we get

$x_{1}+x_{2}=\frac{\sqrt{3}}{4}a^{2}+\frac{\sqrt{3}}{4}b^{2}$

$x_{1}+x_{2}=\frac{\sqrt{3}}{4}\left ( a^{2}+b^{2} \right )=x_{3}$

Hence $x_{1}+x_{2}=x_{3}$

Hence proved

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#6.4

In a quadrilateral ABCD, $\angle A + \angle D = 90^{o}$ . Prove that $AC^{2} + BD^{2} = AD^{2} + BC^{2}$

Given : ABCD is a quadrilateral, $\angle A+\angle D=90^{o}$

To prove :- $AC^{2} + BD^{2} = AD^{2} + BC^{2}$

Proof : In $\Delta ADE$

$\angle A+\angle D=90^{o} \; \; \; \; \; ....(1)$ (given)

for find $\angle E$ use sum of the angle of a triangle is equal to $180^0$

$\angle A+\angle D+\angle E=180^{o}$

$90+\angle E=180^{o}$

$\angle E=180^{o} -90^{o}$

$\angle E=90^{o}$

In $\Delta ADE$ use the Pythagoras theorem we get

$AD^{2}=AE^{2}+DE^{2}\; \; \; \; \; \; \; \; ....(2)$

In $\Delta BEC$ use the Pythagoras theorem we get

$BC^{2}=BE^{2}+EC^{2} ....(3)$

Add equation (2) and (3) we get

$AD^{2}+BC^{2}=AE^{2}+DE^{2}+BE^{2}+CE^{2} .....(4)$

In $\Delta ACE$ use the Pythagoras theorem we get

$AC^{2}=AE^{2}+CE^{2}....(5)$

In $\Delta EBD$ use the Pythagoras theorem we get

$BD^{2}=BE^{2}+DE^{2}\; \; \; \; \; \; \; \; \; \; \; \; ....(6)$

Now add equation (5) and (6) we get

$AC^{2} + BD^{2} = AE^{2} + CE^{2} + BE^{2} + DE^{2}....(7)$

From equation (4) and (7) we get

$AC^{2} + BD^{2} = AD^{2} + BC^{2}$

Hence proved.

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#6.4

$\text{In} \Delta PQR,$ $PD \perp QR$ $\text{such that D lies on QR}$ . $\text{If}$ $PQ = a, PR = b, QD = c$ $\text{and}$ $DR=d$ $\text{, prove that }$ $(a + b) (a - b) = (c + d) (c - d).$

and

$\text{Given:- PQR is a triangle and}$ $PD\perp QR$

$PQ = a, PR = b, QD = C, DR = d$

To prove:-

$(a + b) (a - b) = (c + d) (c - d)$

$\text{Proof :- In}$ $\Delta PQD$ $\text{use Pythagoras theorem}$

$\\PQ^{2}=QD^{2}+PD^{2}$

$\\a^{2}=c^{2}+PD^{2}\\a^{2}-c^{2}=PD^{2} \; \; \; \; \; \; \; \; ....(1)$

$\text{In }$ $\Delta PRD$ $\text{use Pythagoras theorem }$

$\\PR^{2}=PD^{2}+DR^{2}\\b^{2}=PD^{2}+d^{2}\\b^{2}-d^{2}=PD^{2}\; \; \; \; \; \; \; ....(2)$

equate equation (1) and (2) we get
$\\a^{2} - c^{2} = b^{2} - d^{2}\\a^{2}-b^{2}=c^{2}-d^{2}$

$(a - b) (a + b) = (c - d) (c + d)$ $\left [ \because a^{2} - b^{2} = (a - b) (a + b) \right ]$

Hence proved.

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#6.4

In Fig., PQR is a right triangle right angled at Q and $QS \perp PR$ . If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.

Given: PQR is a triangle

$\angle Q = 90^{o} \; and\; QS \perp PR$

PQ = 6 cm, PS = 4 cm

In $\Delta SQP \; and \; \Delta SRQ$

$\angle S=\angle S$ (common angles and each angle is 90°)

$\angle SPQ=\angle SQR$ (each equal to $90^{o}-\angle R$ )

$\therefore \Delta SQP \sim \Delta SRQ$ ( by AA similarity criterion)

$\Rightarrow \frac{SQ}{PS}=\frac{SR}{SQ}$

By cross multiply we get

$SQ^{2}=PS.SR \; \; \; \; \; \; \; ....(1)$

In $\Delta PSQ$ use Pythagoras theorem.

$PQ^{2}=PS^{2}+QS^{2}$

$6^{2}=4^{2}+QS^{2}$

$36-16=QS^{2}$

$20=QS^{2}$

$QS=\sqrt{20}=2\sqrt{5}cm$

Put $QS=2\sqrt{5}$ in equation (1)

$\left ( 2\sqrt{5} \right )^{2}=4 \times SR$

$\frac{20}{4}=SR$

$5cm=SR$

In $\Delta QSR$ use Pythagoras theorem

$QR^{2}=QS^{2}+SR^{2}$

$QR^{2}=\left ( 2\sqrt{5} \right )^{2}+\left ( 5 \right )^{2}$

$QR^{2}=20+25$

$QR=\sqrt{45}=3\sqrt{5}cm$

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#6.4

In Fig., ABC is a triangle right angled at B and $BD \perp AC$ . If AD = 4 cm, and CD = 5 cm, find BD and AB.

Answer: $\left [ 2\sqrt{5}\; cm\; and\; 6\; cm \right ]$

Given :- $\angle B=90^{o}$ and $BD\perp AC$

AD = 4 cm and CD = 5 cm

In $\Delta ABD$ and $\Delta BDC$

$\angle ADB = \angle BDC$ (each equal to 90^o)

$\angle BAD = \angle DBC$ (each equal to 90^o-C)

$\therefore \Delta ABD\sim \Delta BDC$ (by AA similarity criterion)

$\Rightarrow \frac{DB}{DA}=\frac{DC}{DB}$

By cross multiply we get

$DB^{2}=DA.DC$

$DB^{2}=4 \times 5$

$DB=\sqrt{20}=2\sqrt{5}cm$

In $\Delta BDC$ use Pythagoras theorem

$BC^{2}=BD^{2}+CD^{2}$

$BC^{2}=\left ( 2\sqrt{5} \right )^{2}+\left ( 5 \right )^{2}$

$BC^{2}=20+25=45$

$BC=\sqrt{45}=3\sqrt{5}$

We know that $\Delta DBA\sim \Delta DBC$

$\therefore \frac{DB}{DC}=\frac{BA}{BC}\Rightarrow \frac{2\sqrt{5}}{5}=\frac{BA}{3\sqrt{5}}$

$BA=\frac{2\sqrt{5}\times 3\sqrt{5}}{5}=\frac{6 \times 5}{5}=6\; cm$

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In fig. , l || m and line segments AB, CD and EF are concurrent at point P. Prove tha t.

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In Fig., PA, QB, RC and SD are all perpendiculars to a line l, AB = 6 cm, BC = 9 cm, CD = 12 cm and SP = 36 cm. Find PQ, QR and RS.

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O is the point of intersection of the diagonals AC and BD of a trapezium ABCD with AB || DC. Through O, a line segment PQ is drawn parallel to AB meeting AD in P and BC in Q. Prove that PO = QO.

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In Fig., line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and .Prove that .

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JEE Main high-scoring chapters and topics

Prove that the area of the semicircle drawn on the hypotenuse of a right angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

In fig. , l || m and line segments AB, CD and EF are concurrent at point P. Prove tha

t $\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$ .

In Fig., line segment DF intersect the side AC of a triangle ABC at the point E such that E is the mid-point of CA and $\angle AEF=\angle AFE$ .Prove that $\frac{BD}{CD}=\frac{BF}{CE}$ .

In a quadrilateral ABCD, $\angle A + \angle D = 90^{o}$ . Prove that $AC^{2} + BD^{2} = AD^{2} + BC^{2}$

$\text{In} \Delta PQR,$ $PD \perp QR$ $\text{such that D lies on QR}$ . $\text{If}$ $PQ = a, PR = b, QD = c$ $\text{and}$ $DR=d$ $\text{, prove that }$ $(a + b) (a - b) = (c + d) (c - d).$

and

In Fig., PQR is a right triangle right angled at Q and $QS \perp PR$ . If PQ = 6 cm and PS = 4 cm, find QS, RS and QR.

In Fig., ABC is a triangle right angled at B and $BD \perp AC$ . If AD = 4 cm, and CD = 5 cm, find BD and AB.