Chapter 6: Triangles

Ans: It is given that $\mathrm{ABC}$ is an isosceles triangle.

$\therefore \mathrm{AB}=\mathrm{AC}$

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

In $\Delta \mathrm{ABD}$ and $\triangle \mathrm{ECF}$

$\angle \mathrm{ADB}=\angle \mathrm{EFC}\left(\operatorname{Each} 90^{\circ}\right)$

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

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

Ans: Median equally divides the opposite side.

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

Given that,

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

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

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

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

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \text { (Proved above) }$

$\therefore \Delta \mathrm{ABD} \sim \Delta \mathrm{PQM}$ (By SSS similarity criterion)

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

In $\triangle \mathrm{ABC}$ and $\Delta \mathrm{PQR}$,

$\angle \mathrm{ABD}=\angle \mathrm{PQM} \text { (Proved above) }$

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

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$ (By SAS similarity criterion)

Ans: $\text { In } \triangle \mathrm{ADC} \text { and } \triangle \mathrm{BAC} \text {, }$

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

$\therefore \Delta \mathrm{ADC} \sim \triangle \mathrm{BAC}$ (By AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion. $\therefore \frac{\mathrm{CA}}{\mathrm{CB}}=\frac{\mathrm{CD}}{\mathrm{CA}}$ $\Rightarrow \mathrm{CA}^{2}=\mathrm{CB} \cdot \mathrm{CD}$

Ans: Given that,

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

Let us extend AD and PM up to point $E$ and $L$ respectively, such that $A D=$ 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, $\mathrm{BD}=\mathrm{DC}$ and $\mathrm{QM}=\mathrm{MR}$

Also, $\mathrm{AD}=\mathrm{DE}$ (By construction)

And, $\mathrm{PM}=\mathrm{ML}$ (By construction)

In quadrilateral ABEC, diagonals AE and BC bisect each other at point D. Therefore, quadrilateral ABEC is a parallelogram.

$\therefore \mathrm{AC}=\mathrm{BE}$ and $\mathrm{AB}=\mathrm{EC}$ (Opposite sides of a parallelogram are equal) Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, $P Q=L R$

It was given that

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

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

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

$\therefore \triangle \mathrm{ABE} \sim \triangle \mathrm{PQL}$ (By SSS similarity criterion)

We know that corresponding angles of similar triangles are equal.

$\therefore \angle \mathrm{BAE}=\angle \mathrm{QPL} \ldots$ (1)

Similarly, it can be proved that $\triangle \mathrm{AEC} \sim \triangle \mathrm{PLR}$ and

$\angle \mathrm{CAE}=\angle \mathrm{RPL} \ldots$ (2)

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

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

$\Rightarrow \angle \mathrm{CAB}=\angle \mathrm{RPQ} \ldots$ (3)

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

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

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

$\therefore \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$ (By SAS similarity criterion)

Ans: Let $\mathrm{AB}$ and $\mathrm{CD}$ be a tower and a pole respectively.

Let the shadow of $\mathrm{BE}$ and DF be the shadow of $\mathrm{AB}$ and $\mathrm{CD}$ respectively.

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

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

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

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

$\therefore \Delta \mathrm{ABE} \sim \Delta \mathrm{CDF}$ (AAA similarity criterion)

$\Rightarrow \frac{\mathrm{AB}}{\mathrm{CD}}=\frac{\mathrm{BE}}{\mathrm{DF}}$

$\Rightarrow \frac{\mathrm{AB}}{6 \mathrm{~cm}}=\frac{28}{4}$

$\Rightarrow \mathrm{AB}=42 \mathrm{~m}$

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

Ans: It is given that $\triangle \mathrm{ABC} \sim \Delta \mathrm{PQR}$

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

$\therefore \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}} \ldots(1)$

Also, $\angle \mathrm{A}=\angle \mathrm{P}, \angle \mathrm{B}=\angle \mathrm{Q}, \angle \mathrm{C}=\angle \mathrm{R} \quad \ldots$ (2)

Since AD and PM are medians, they will divide their opposite sides. $\therefore \mathrm{BD}=\frac{\mathrm{BC}}{2}$ and $\mathrm{QM}=\frac{\mathrm{QR}}{2}$

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

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}$

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

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

$\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}$

(Using equation (4))

$\therefore \triangle \mathrm{ABD} \sim \triangle \mathrm{PQM}$ (By SAS similarity criterion)

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