Question 11: In $\triangle PQR, \angle Q=90^o$ and $QM \perp PR$. Prove that:

(i) $PQ^2 = PM \times PR$

(ii) $QR^2 = PR \times MR$

(iii) $PQ^2 + QR^2 = PR^2$

(i)  Given  $\triangle PQR, \angle Q=90^o$ and $QM \perp PR$

Consider $\triangle PQM \ and\ \triangle PRQ$

$\angle PMQ = \angle PQR = 90^o$ (given)

$\angle QPM =\angle RPQ$ (common angle)

Therefore $\triangle PQM \sim \triangle PRQ$   (AAA postulate)

Hence $\frac{PQ}{PR} = \frac{PM}{PQ}$

Therefore  $PQ^2 = PM \times PR$

(ii) Consider $\triangle QMR \ and\ \triangle PQR$

$\angle QMR = \angle PQR = 90^o$ (given)

$\angle QRM =\angle QRP$ (common angle)

Therefore $\triangle QMR \sim \triangle PQR$   (AAA postulate)

Hence $\frac{QR}{PR} = \frac{MR}{QR}$

Therefore  $QR^2 = PR \times MR$

$PQ^2 + QR^2 = PM \times PR + PR \times MR$

$\Rightarrow PQ^2 + QR^2 = PR \times (PM + MR)$

$\Rightarrow PQ^2 + QR^2 = PR \times PR = PR^2$

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Question 12: In $\triangle ABC, \angle C= 90^o, CD \perp AB$. Prove that $CD^2 = AD \times DB$.

$\triangle ABC, \angle C= 90^o, CD \perp AB$.

Consider $\triangle PQM \ and\ \triangle PRQ$

$\angle PMQ = \angle PQR = 90^o$ (given)

$\angle QPM =\angle RPQ$ (common angle)

Therefore $\triangle PQM \sim \triangle PRQ$   (AAA postulate)

Hence $\frac{PQ}{PR} = \frac{PM}{PQ}$

Therefore  $PQ^2 = PM \times PR$

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Question 13: In $\triangle ABC, \angle B= 90^o, BD \perp AC$.

(i) If $CD=10 \ cm \ and \ BD = 8 \ cm, \ find \ AD$

(ii) If $AC=18 \ cm \ and \ AD = 6 \ cm, \ find \ BD$

(iii) If $AC=9 \ cm \ and \ AB = 7 \ cm, \ find \ AD$

$\triangle ABC, \angle B= 90^o, BD \perp AC$

Consider $\triangle CDB \ and\ \triangle BCD$

$\angle ADB = \angle BDC = 90^o$ (given)

$\angle ABD =\angle BCD$

Therefore $\triangle CDB \sim \triangle BDA$   (AAA postulate)

(i)   $\frac{CD}{BD} = \frac{BD}{AD}$

Therefore $AD = \frac{BD^2}{CD} = \frac{8^2}{10} = 6.4\ cm$

(ii)  $\frac{BD}{DA} = \frac{CD}{BD}$

Therefore $BD = \sqrt{CD \times DA} = \sqrt{72}= 8.49 \ cm$

(iii)  $\frac{AD}{AB} = \frac{AB}{AC}$

Therefore $AD = \frac{AB \times AB}{AC} = \frac{7 \times 7}{9} = 5.44 \ cm$

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Question 14: In the figure, $PQRS$ is a parallelogram with $PQ=16 \ cm$ and $QR=10 \ cm$. $L$ is a point on $PR$ such that $RL:LP=2:3$. $QL$ produced meets $RS$ at $M$ and $PS$ produced at $N$.  Find the lengths of $PN$ and $RM$ [1997]

In $\triangle RLQ$ and $\triangle PLN$

$\angle RLQ = \angle PLN$ (vertically opposite angles)

$\angle LRQ = \angle LPN$ (alternate angles)

Therefore $\triangle RLQ \sim \triangle PLN$ (AAA postulate)

Therefore $\frac{RL}{LP}=\frac{RQ}{PN}$

$\Rightarrow PN = \frac{10}{2} \times 3 = 15 \ cm$

In $\triangle RLM \ and \ \triangle PLQ$

$\angle RLM = \angle PLQ$ (vertically opposite angles)

$\angle LRM = \angle LPQ$ (alternate angles)

Therefore $\triangle RLM \sim \triangle PLQ$ (AAA postulate)

Therefore $\frac{RM}{PQ}=\frac{RL}{LP}$

$\Rightarrow RM = \frac{2}{3} \times 16 = 10.667 \ cm$

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Question 15:  In quadrilateral $ABCD$, Diagonal $AC \ and \ BD$ intersect at point $E$ such that: $AE:EC=BE:ED$. Show that $ABCD$ is a parallelogram.

Given $\frac{AE}{EC} = \frac{BE}{ED}$ … … … … (i)

In $\triangle ADB$, because $EF \parallel AB$, by proportionality theorem

$\frac{DF}{FA} = \frac{ED}{BE}$ … … … … (i)

Therefore  from (i) and (ii)

$\frac{DF}{FA} = \frac{EC}{AE}$

Similarly in $\triangle DCA$

$\frac{DF}{FA} = \frac{CE}{AE} \Rightarrow FE \parallel DC$

Also it is given that $FE \parallel AB$

Therefore $DC \parallel AB$. Therefore $ABCD$ is a parallelogram.

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Question 16: Given $AB \parallel DE$ and $BC \parallel EF$. Prove that:

(i) $\frac{AD}{DG}=\frac{CF}{FG}$

(ii) $\triangle DFG \sim \triangle ACG$

(i) Given $AB \parallel DE$ and $BC \parallel EF$

In $\triangle AGB, \frac{AD}{DG}=\frac{EB}{GE}$… … … … (i)

In $\triangle BGC, \frac{EB}{GE}=\frac{CF}{FG}$… … … … (ii)

From (i) and (ii)  you get $\frac{AD}{DG}=\frac{CF}{FG}$

(ii)  Consider $\triangle DFG \ and\ \triangle ACG$

We have $\frac{AD}{DG}=\frac{CF}{FG}$

and $\angle DGF = \angle AGC$ (common angle)

Therefore  $\triangle DFG \sim \triangle ACG$

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Question 17: In $\triangle ABC, AD \perp BC$ and $AD^2 = BD \times DC$. Show that $\angle BAC = 90^o$.

Given $AD^2 = BD \times DC$

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

Also given $\angle ADB = \angle ADC = 90^o$

Therefore $\triangle DBA \sim \triangle DAC$ (SAS postulate)

Therefore $\angle C = \angle BAD$

$\angle B = \angle DAC$

Adding $\angle C + \angle B = \angle BAD + \angle DAC$

$\Rightarrow \angle A =\angle C + \angle B$ … … … (i)

We know $\angle A + \angle B + \angle C = 180^o$ … … … … (ii)

From (i) and (ii) $2 \angle A = 180^o$

$\Rightarrow \angle A = 90^o$

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Question 18:  In the given figure $AB \parallel EF \parallel DC$; $AB=67.5 \ cm, DC = 40.5 \ cm$ and $AE = 52.5 \ cm$.

(i) Name the three pairs of similar triangles

(ii) Find the lengths of $EC \ and \ EF$

Given $AB \parallel EF \parallel DC$,  The three pairs of similar triangles are

$\triangle ABC \sim \triangle CEF$

$\triangle BCD \sim \triangle BEF$

$\triangle ABE \sim \triangle CDE$

Since $\triangle ABE \sim \triangle CDE$

$\frac{AB}{CD}=\frac{AE}{CE}$ $\Rightarrow CE =$ $\frac{52.5 \times 40.5}{67.5}$ $= 31.5 \ cm$

Since $\triangle ABC \sim \triangle CEF$

$\frac{CE}{CA}=\frac{EF}{AB}$ $\Rightarrow EF =$ $\frac{67.5 \times 31.5}{52.5+31.5} =$ $25.3125 \ cm$

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Question 19: In the given figure $QR \parallel AB$, and $DR \parallel QB$.  Prove that $PQ^2 = PD \times PA$.

Given $QR \parallel AB$, and $DR \parallel QB$

Using basic proportionality theorem

In $\triangle PQR \ and \ \triangle PAB$

$\frac{PQ}{PA}=\frac{PR}{PB}$   … … … … (i)

In $\triangle PDR \ and \ \triangle PQB$

$\frac{PD}{PQ}=\frac{PR}{PB}$   … … … … (ii)

From (i) and (ii)  we get

$\frac{PQ}{PA} = \frac{PD}{PQ}$

$\Rightarrow PQ^2 = PD \times PA$

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Question 20: Through the mid point $M$ of the side $CD$ of a parallelogram $ABCD$, the line $BM$ is drawn intersecting diagonal $AC \ in \ L$ and $AD$ produced in $E$. Prove that $EL=2BL$.

Consider $\triangle DEM \ and \ \triangle CBM$

$\angle AEB = \angle EBC$ (alternate angles)

$\angle EMD = \angle CMB$ (opposite angles)

$DM = MC$ (as M is the midpoint of CD)

Therefore $\triangle DEM \cong \triangle CBM$

Therefore $DE = BC$ (corresponding angles)

$AD = BC$ (opposite sides of a parallelogram)

Therefore $AE = AD + DE = 2BC$

Now consider $\triangle ELA \ and \ \triangle LBC$

$\angle AEB = \angle EBC$

$\angle ELA = \angle EBC$

Therefore $\triangle ELA \sim \triangle LBC$

Hence $\frac{EL}{BL}= \frac{EA}{BC}$

$\frac{EL}{BL}= \frac{2BC}{BC} = 2$

Hence $EL = 2 BL$

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Question 21: In the figure given below,  $P$ is a point on $AB$ such that $AP:PB=4:3$. $PQ \parallel AC$.

(i) Calculate the ratio $PQ:AC$, giving reasons for your answer.

(ii) In $\triangle ARC, \angle ARC = 90^o$ and in $\triangle PQS, \angle PSQ=90^o$. Given $QS=6 \ cm$, calculate length of $AR$  [1999]

(i)   Given $AP:PB=4:3$

Also $PQ \parallel AC$, applying basic proportionality theorem

$\frac{AP}{PB}=\frac{CQ}{QB}$

$\Rightarrow \frac{CQ}{QB}= \frac{4}{3}$

$\Rightarrow \frac{BQ}{BC}=\frac{3}{7}$

$\angle QBP = \angle ACB$ (corresponding angles)

$\angle QPB = \angle CAB$ (corresponding angles)

Therefore $\triangle PBQ \sim \triangle ABC$ (AAA postulate)

$\frac{PQ}{ AC} = \frac{BQ}{ BC} = \frac{3}{ 7}$

(ii) Given $\angle ARC = \angle QSP = 90^o$

$\angle ACR = \angle SPQ$ (alternate angles)

Therefore $\triangle ARC \sim \triangle QSP$ (AAA postulate)

$\frac{AR}{ SQ} = \frac{AC}{ PQ} = \frac{7}{ 3}$

$\Rightarrow AR = \frac{7 \times 6}{3} = 14 \ cm$

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Question 22: In the right angled $\triangle QPR, PM$ is the altitude. Given that $QR= 8 cm$ and $MQ = 3.5 cm$, calculate the value of $PR$. [2000]

$\angle QPR = \angle PMR = 90^o$

$\angle PRQ = \angle PRM$ (common)

$\triangle PQR \sim \triangle MPR$ (AAA postulate)

$\frac{QR}{ PR}= \frac{ PR}{ MR}$

$PR^2 = QR \times MR = 8 \times 4.5 = 36 \Rightarrow PR = 6 \ cm$

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Question 23: In the figure given below, median $BD \ and \ CE$ of the $\triangle ABC$ meet at $G$. Prove that:

(i) $\triangle EGD \sim \triangle CGB$

(ii) $BG = 2 GD$ from (i)

Given $BD \ and \ CE$ are medians

Therefore $AD = DC$ and $AE = BE$

Applying converse of proportionality theorem $ED \parallel BC$

In $\triangle EGD$ and $\triangle CGB$

$\angle DEG = \angle GCB$ (alternate angles)

$\angle EGD = \angle BGC$ (vertically opposite angles)

Therefore $\triangle EGD \sim \triangle CGB$ (AAA postulate)

Therefore $\frac{GD}{GB}=\frac{ED}{BC}$

In $\triangle EAD$ and $\triangle BAC$

$\angle AED = \angle ABC$ (corresponding angles)

$\angle EAD = \angle BAC$ (common angle)

Therefore $\triangle EAD \sim \triangle BAC$ (AAA postulate)

Therefore $\frac{ED}{BC}=\frac{AE}{AB} = \frac{1}{2}$

$\frac{ED}{BC} = \frac{1}{2}$

From 1,  $\frac{GD}{GB} = \frac{1}{2}$

$GB = 2GD$

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