Solution

(i) $\mathrm{\angle }A=\mathrm{\angle }P={60}^{\circ }$

$\mathrm{\angle }B=\mathrm{\angle }Q={80}^{\circ }$

$\mathrm{\angle }C=\mathrm{\angle }R={40}^{\circ }$

Therefore, $\mathrm{△}ABC\tilde{A}\varphi \ddot{E}†\hat{A}1{\displaystyle \frac{1}{4}}\mathrm{\Delta }PQR$ [By AAA similarity criterion]

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

(ii)

$\therefore \mathrm{△}ABC\sim \mathrm{△}QRP$

[By SSS similarity criterion]

(iii)The given triangles are not similar as the corresponding sides are not proportional.

Solution

DOB is a straight line.

$\therefore \mathrm{\angle }DOC+\mathrm{\angle }COB={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DOC={180}^{\circ }-{125}^{\circ }$

$={55}^{\circ }$

In $\mathrm{△}DOC$,

$\mathrm{\angle }DCO+\mathrm{\angle }CDO+\mathrm{\angle }DOC={180}^{\circ }$

(Sum of the measures of the angles of a triangle is ${180}^{\circ }$.)

$\Rightarrow \mathrm{\angle }DCO+{70}^{\circ }+{55}^{\circ }={180}^{\circ }$

$\Rightarrow \mathrm{\angle }DCO={55}^{\circ }$

It is given that $\mathrm{△}ODC\mathrm{\angle }\hat{A}1/4\mathrm{△}OBA$.

$\therefore \mathrm{\angle }OAB=\mathrm{\angle }OCD$ [Corresponding angles are equal in similar triangles.]

$\Rightarrow \mathrm{\angle }OAB={55}^{\circ }$

Solution

In $\mathrm{△}PQR,\mathrm{\angle }PQR=\mathrm{\angle }PRQ$

$\therefore PQ=PR(i)$

Given, ${\displaystyle \frac{QR}{QS}}={\displaystyle \frac{QT}{PR}}$

Using $(i)$, we obtain

${\displaystyle \frac{QR}{QS}}={\displaystyle \frac{QT}{QP}}$

In $\mathrm{△}PQS$ and $\mathrm{△}TQR$.

$\begin{array}{rl}& {\displaystyle \frac{QR}{QS}}={\displaystyle \frac{QT}{QP}}[\text{ Using }(ii)]\\ & \mathrm{\angle }Q=\mathrm{\angle }Q\\ & \therefore \mathrm{△}PQS\sim \mathrm{△}TQR\text{ [SAS similarity criterion] }\end{array}$

Solution

In $\mathrm{△}RPQ$ and $\mathrm{△}RST$,

$\mathrm{\angle }RTS=\mathrm{\angle }QPS$ (Given)

$\mathrm{\angle }R=\mathrm{\angle }R$ (Common angle)

$\therefore \mathrm{\Delta }RPQ\propto 1/4\mathrm{\Delta }RTS$ (By AA similarity criterion)

Solution

It is given that $\mathrm{△}ABE\cong \mathrm{△}ACD$.

$\therefore \text{AB=AC[By CPCT] (1) }$

And, $\text{AD=AE[By CPCT]}(2)$

In $\mathrm{△}ADE$ and $\mathrm{△}ABC$,

${\displaystyle \frac{AD}{AB}}={\displaystyle \frac{AE}{AC}}$

[Dividing equation (2) by (1)]

$\mathrm{\angle }A=\mathrm{\angle }A[$ Common angle]

$\therefore \mathrm{△}ADE\sim \mathrm{\Delta }ABC$ [By SAS similarity criterion]

Solution

(i)

In $\mathrm{△}AEP$ and $\mathrm{△}CDP$,

$\mathrm{\angle }AEP=\mathrm{\angle }CDP($ Each 90 $)$

$\mathrm{\angle }APE=\mathrm{\angle }CPD$ (Vertically opposite angles)

Hence, by using AA similarity criterion,

$\mathrm{△}AEP\propto 1/4\mathrm{\Delta }CDP$

(ii)

In $\mathrm{△}ABD$ and $\mathrm{△}CBE$,

$\mathrm{\angle }ADB=\mathrm{\angle }CEB(.$ Each ${.90}^{\circ })$

$\mathrm{\angle }ABD=\mathrm{\angle }CBE$ (Common)

Hence, by using AA similarity criterion,

$\mathrm{△}ABD\propto 1/4\mathrm{△}CBE$

(iii)

In $\mathrm{△}AEP$ and $\mathrm{△}ADB$,

$\mathrm{\angle }AEP=\mathrm{\angle }ADB(.$ Each ${.90}^{\circ })$

$\mathrm{\angle }PAE=\mathrm{\angle }DAB$ (Common)

Hence, by using AA similarity criterion,

$\mathrm{△}AEP\propto 1/4\mathrm{\Delta }ADB$

(iv)

In $\mathrm{△}PDC$ and $\mathrm{△}BEC$,

$\mathrm{\angle }PDC=\mathrm{\angle }BEC(.$ Each ${.90}^{\circ })$

$\mathrm{\angle }PCD=\mathrm{\angle }BCE$ (Common angle)

Hence, by using AA similarity criterion,

$\mathrm{△}PDC\propto 1/4\mathrm{△}BEC$

Solution

In $\mathrm{△}ABE$ and $\mathrm{△}CFB$,

$\mathrm{\angle }A=\mathrm{\angle }C$ (Opposite angles of a parallelogram)

$\mathrm{\angle }AEB=\mathrm{\angle }CBF$ (Alternate interior angles as $AE\Vert BC$ )

$\therefore \mathrm{△}ABE\propto 1/4\mathrm{△}CFB$ (By AA similarity criterion)

Solution

In $\mathrm{△}ABC$ and $\mathrm{△}AMP$,

$\mathrm{\angle }ABC=\mathrm{\angle }AMP(.$ Each ${.90}^{\circ })$

$\mathrm{\angle }A=\mathrm{\angle }A$ (Common)

$\therefore \mathrm{△}ABC\tilde{A}\varphi \ddot{E}+\hat{A}1/4\mathrm{\Delta }AMP$ (By AA similarity criterion)

$\Rightarrow {\displaystyle \frac{CA}{PA}}={\displaystyle \frac{BC}{MP}}$ (Corresponding sides of similar triangles are proportional)

Solution

Answer :

It is given that $\mathrm{△}ABC\tilde{A}\varphi \ddot{E}†\hat{A}114\mathrm{\Delta }FEG$.

$\therefore \mathrm{\angle }A=\mathrm{\angle }F,\mathrm{\angle }B=\mathrm{\angle }E$, and $\mathrm{\angle }ACB=\mathrm{\angle }FGE$ $\mathrm{\angle }ACB=\mathrm{\angle }FGE$

$\therefore \mathrm{\angle }ACD=\mathrm{\angle }FGH$ (Angle bisector)

And, $\mathrm{\angle }DCB=\mathrm{\angle }HGE$ (Angle bisector)

In $\mathrm{△}ACD$ and $\mathrm{△}FGH$,

$\mathrm{\angle }A=\mathrm{\angle }F$ (Proved above)

$\mathrm{\angle }ACD=\mathrm{\angle }FGH$ (Proved above)

$\therefore \mathrm{△}ACD\sim \mathrm{\Delta }FGH$ (By AA similarity criterion)

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

In $\mathrm{△}DCB$ and $\mathrm{△}HGE$,

$\mathrm{\angle }DCB=\mathrm{\angle }HGE$ (Proved above)

$\mathrm{\angle }B=\mathrm{\angle }E$ (Proved above)

$\therefore \mathrm{△}DCB\tilde{A}\varphi \ddot{E}+A^1/4\mathrm{\Delta }HGE$ (By AA similarity criterion)

In $\mathrm{△}DCA$ and $\mathrm{△}HGF$,

$\mathrm{\angle }ACD=\mathrm{\angle }FGH$ (Proved above)

$\mathrm{\angle }A=\mathrm{\angle }F$ (Proved above)

$\therefore \mathrm{△}DCA\sim \mathrm{\Delta }HGF$ (By AA similarity criterion)

Solution

It is given that $ABC$ is an isosceles triangle.

$\therefore AB=AC$

$\Rightarrow \mathrm{\angle }ABD=\mathrm{\angle }ECF$

In $\mathrm{△}ABD$ and $\mathrm{△}ECF$, $\mathrm{\angle }ADB=\mathrm{\angle }EFC(.$ Each ${.90}^{\circ })$

$\mathrm{\angle }BAD=\mathrm{\angle }CEF$ (Proved above)

$\therefore \mathrm{\Delta }ABD\sim \mathrm{\Delta }ECF$ (By using $AA$ similarity criterion)

Solution

Median divides the opposite side.

$\therefore BD={\displaystyle \frac{BC}{2}}\text{ and }QM={\displaystyle \frac{QR}{2}}$

Given that,

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

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

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

In $\mathrm{△}ABD$ and $\mathrm{△}PQM$,

$\begin{array}{rl}& {\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{BD}{QM}}={\displaystyle \frac{AD}{PM}}\\ & \therefore \mathrm{△}ABD\tilde{A}\varphi \ddot{E}†\hat{A}1/4\mathrm{\Delta }PQM\text{ (By SSS similarity criterion) }\\ & \Rightarrow \mathrm{\angle }ABD=\mathrm{\angle }PQM\text{ (Corresponding angles of similar triangles) }\end{array}$

In $\mathrm{△}ABC$ and $\mathrm{△}PQR$,

$\mathrm{\angle }ABD=\mathrm{\angle }PQM$ (Proved above) ${\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{BC}{QR}}$

$\therefore \mathrm{△}ABC\tilde{A}\varphi \ddot{E}+\hat{A}11/4\mathrm{△}PQR$ (By SAS similarity criterion)

Solution

In $\mathrm{△}ADC$ and $\mathrm{△}BAC$,

$\mathrm{\angle }ADC=\mathrm{\angle }BAC$ (Given)

$\mathrm{\angle }ACD=\mathrm{\angle }BCA$ (Common angle)

$\therefore \mathrm{△}ADC\tilde{A}\varphi \ddot{E}+\hat{A}11/4\mathrm{\Delta }BAC$ (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

$\therefore {\displaystyle \frac{CA}{CB}}={\displaystyle \frac{CD}{CA}}$

$\Rightarrow C{A}^2=CB\times CD$

Solution

Given that,

${\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{AC}{PR}}={\displaystyle \frac{AD}{PM}}$

Let us extend $AD$ and $PM$ up to point $E$ and $L$ respectively, such that $AD=DE$ and $PM=ML$. Then, join $B$ to $E,C$ to $E,Q$ to $L$, and $R$ to $L$.

We know that medians divide opposite sides.

Therefore, $BD=DC$ and $QM=MR$

Also, $AD=DE$ (By construction)

And, $PM=ML$ (By construction)

In quadrilateral $ABEC$, diagonals $AE$ and $BC$ bisect each other at point $D$.

Therefore, quadrilateral ABEC is a parallelogram.

$\therefore AC=BE$ and $AB=EC$ (Opposite sides of a parallelogram are equal)

Similarly, we can prove that quadrilateral $PQLR$ is a parallelogram and $PR=QL,PQ=LR$

It was given that

$\begin{array}{rl}& {\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{AC}{PR}}={\displaystyle \frac{AD}{PM}}\\ & \Rightarrow {\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{BE}{QL}}={\displaystyle \frac{2AD}{2PM}}\\ & \Rightarrow {\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{BE}{QL}}={\displaystyle \frac{AE}{PL}}\end{array}$

$\therefore \mathrm{△}ABE\tilde{A}\varphi \ddot{E}†\hat{A}1/4\mathrm{\Delta }PQL$ (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

$\therefore \mathrm{\angle }BAE=\mathrm{\angle }QPL$.

Similarly, it can be proved that $\mathrm{△}AEC\tilde{A}\varphi \ddot{E}†\hat{A}11/4\mathrm{\Delta }PLR$ and

$\mathrm{\angle }CAE=\mathrm{\angle }RPL\dots$

Adding equation (1) and (2), we obtain

$\mathrm{\angle }BAE+\mathrm{\angle }CAE=\mathrm{\angle }QPL+\mathrm{\angle }RPL$

$\Rightarrow \mathrm{\angle }CAB=\mathrm{\angle }RPQ$

In $\mathrm{△}ABC$ and $\mathrm{△}PQR$,

${\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{AC}{PR}}$

(Given)

$\mathrm{\angle }CAB=\mathrm{\angle }RPQ$ [Using equation (3)]

$\therefore \mathrm{△}ABC\tilde{A}\varphi \ddot{E}†\hat{A}1{\displaystyle \frac{1}{4}}\mathrm{\Delta }PQR$ (By SAS similarity criterion)

Solution

Let $AB$ and $CD$ be a tower and a pole respectively.

Let the shadow of $BE$ and $DF$ be the shadow of $AB$ and $CD$ respectively.

At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.

Therefore, $\mathrm{\angle }DCF=\mathrm{\angle }BAE$

And, $\mathrm{\angle }DFC=\mathrm{\angle }BEA$

$\mathrm{\angle }CDF=\mathrm{\angle }ABE$ (Tower and pole are vertical to the ground)

$\therefore \mathrm{△}ABE\sim \mathrm{\Delta }CDF$ (AAA similarity criterion) $\Rightarrow {\displaystyle \frac{AB}{CD}}={\displaystyle \frac{BE}{DF}}$

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

$\Rightarrow AB=42m$

Therefore, the height of the tower will be 42 metres.

Solution

It is given that $\mathrm{△}ABC\tilde{A}\varphi \ddot{E}†\hat{A}1/4\mathrm{△}PQR$

We know that the corresponding sides of similar triangles are in proportion.

$\therefore {\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{AC}{PR}}={\displaystyle \frac{BC}{QR}}$

Also, $\mathrm{\angle }A=\mathrm{\angle }P,\mathrm{\angle }B=\mathrm{\angle }Q,\mathrm{\angle }C=\mathrm{\angle }R$

Since AD and PM are medians, they will divide their opposite sides.

$$\begin{array}{}\text{(3)}& \therefore BD={\displaystyle \frac{BC}{2}}\text{ and }QM={\displaystyle \frac{QR}{2}}\end{array}$$

From equations ( 1 ) and (3), we obtain

$$\begin{array}{}\text{(4)}& {\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{BD}{QM}}\end{array}$$

In $\mathrm{△}ABD$ and $\mathrm{△}PQM$,

$\mathrm{\angle }B=\mathrm{\angle }Q[$ Using equation (2)] ${\displaystyle \frac{AB}{PQ}}={\displaystyle \frac{BD}{QM}}$

$\therefore \mathrm{△}ABD\tilde{A}\varphi \ddot{E}+\hat{A}11/4\mathrm{△}PQM$ (By SAS similarity criterion)

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