Chapter 6: Triangles

Ans: 

1. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{PQR}$

$\angle \mathrm{A}=\angle \mathrm{P}$

$\angle \mathrm{B}=\angle \mathrm{Q}$

$\angle \mathrm{C}=\angle \mathrm{R}$

$\therefore$ By AAA criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{PQR}$

1. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{QRP}$

$\frac{\mathrm{AB}}{\mathrm{QR}}=\frac{\mathrm{BC}}{\mathrm{RP}}=\frac{\mathrm{AC}}{\mathrm{QP}}=\frac{1}{2}$

$\therefore$ By SSS criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{QRP}$

1. In $\triangle \mathrm{LMP}$ and $\triangle \mathrm{DEF}$

$\frac{\mathrm{LM}}{\mathrm{DE}}=\frac{2.7}{4}, \frac{\mathrm{LP}}{\mathrm{DF}}=\frac{1}{2}$

The sides are not in the equal ratios, Hence the two triangles are not similar.

1. In $\triangle \mathrm{MNL}$ and $\triangle \mathrm{QPR}$

$\angle \mathrm{M}=\angle \mathrm{Q}$

$\frac{\mathrm{MN}}{\mathrm{QP}}=\frac{\mathrm{ML}}{\mathrm{QR}}=\frac{1}{2}$

$\therefore$ By SAS criterion of similarity, $\triangle \mathrm{MNL} \sim \triangle \mathrm{QP} \mathrm{R}$

1. In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{EFD}$

$\angle \mathrm{A}=\angle \mathrm{F}, $

$\frac{AB}{FD}=\frac{BC}{FD}=\frac{1}{2}$

$\therefore$ By SAS criterion of similarity, $\triangle \mathrm{ABC} \sim \triangle \mathrm{EFD}$

1. In $\triangle \mathrm{DEF}$ and $\triangle \mathrm{PQR}$

Since, sum of angles of a triangle is $180^{\circ}$, Hence, $\angle \mathrm{F}=30^{\circ}$ and $\angle \mathrm{P}=70^{\circ}$

$\angle \mathrm{D} =\angle \mathrm{P}$

$\angle \mathrm{E} =\angle \mathrm{Q}$

$\angle \mathrm{F} =\angle \mathrm{R}$

$\therefore$ By AAA criterion of similarity, $\triangle \mathrm{DEF} \sim \triangle \mathrm{PQR}$

Ans: Given:

$\triangle \mathrm{ODC} \sim \triangle \mathrm{OBA}$

$\angle \mathrm{BOC}=125^{\circ}$

$\angle \mathrm{CDO}=70^{\circ}$

To find: $\angle D O C, \angle D C O$ and $\angle O A B$

Sol: Here, $B D$ is a line,

So, we can apply a linear pair on it.

$\angle B O C+\angle D O C=180^{\circ} (Linear Pair)$

$125^{\circ}+\angle D O C=180^{\circ}$

$\angle D O C=180^{\circ}-125^{\circ}$

$\angle D O C=180^{\circ}-125^{\circ}$

$\angle D O C=55^{\circ}$

Now in $\triangle \mathrm{DCO}$

$\angle C D O+\angle D C O+\angle D O C=180^{\circ}$

$70^{\circ}+\angle D C O+55^{\circ}=180^{\circ}$

$125^{\circ}+\angle D C O=180^{\circ}$

$\angle D C O=180^{\circ}-125^{\circ}$

$\angle D C O=55^{\circ}$

Now it is given that

$\triangle O D C \sim \triangle O B A$

Hence,

$\angle \mathrm{DCO}=\angle \mathrm{OAB}$

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

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

Now in $\triangle \mathrm{DCO}$

$\angle \mathrm{CDO}+\angle \mathrm{DCO}+\angle \mathrm{DOC}=180^{\circ} \quad$ (Sum of all angles of triangle is $180^0$ 

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

$125^{\circ}+\angle \mathrm{DCO}=180^{\circ}$ 

$\angle \mathrm{DCO}=180^{\circ}-125^{\circ}$ $\angle \mathrm{DCO}=55^{\circ}$

Now, it is given that

$\Delta \mathrm{ODC} \sim \triangle \mathrm{OBA}$

Hence.

$\angle D C O=\angle O A B$(Corresponding angles of a similar triangles are equal)

$55^{\circ}=\angle O A B$

Ans: In $\Delta \mathrm{DOC}$ and $\triangle \mathrm{BOA}$

$\angle C D O=\angle A B O$ (Alternate interior angles as $A B \| C D)$

$\angle \mathrm{DCO}=\angle \mathrm{BAO}$ (Alternate interior angles as $\mathrm{AB} \| \mathrm{CD})$

$\angle \mathrm{DOC}=\angle \mathrm{BOA}$ (Vertically opposite angles $)$

$\therefore \Delta \mathrm{DOC} \sim \Delta \mathrm{BOA}$ (AAA similarity criterion)

$\therefore \frac{\mathrm{DO}}{\mathrm{BO}}=\frac{\mathrm{OC}}{\mathrm{OA}} \quad($ Corresponding sides are proportional)

$\Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}$

Ans: In $\delta \mathrm{PQR}, \angle \mathrm{PQR}=\angle \mathrm{PRQ}$

$\therefore \mathrm{PQ}=\mathrm{PR}(\mathrm{i})$

Given 

$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}$

Using (i), we obtain

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

In $\triangle \mathrm{PQS}$ and $\triangle \mathrm{TQR}$,

$\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{QP}}[\text { [Using (ii) }]$

$\angle \mathrm{Q}=\angle \mathrm{Q}$

$\therefore \Delta \mathrm{PQS} \sim \Delta \mathrm{TQR} \quad[\text { SAS similarity criterion }]$

Ans: Given: $\Delta P Q R$

and the points $S$ and $T$ on sides PR and QR.

Such that $\angle P=\angle R T S$

To Prove: $\triangle \mathrm{RPQ} \sim \Delta$ RTS.

Proof:

In $\triangle \mathrm{RPQ}$ and $\triangle \mathrm{RTS}$.

$\angle P=\angle R T S$

(Given)

And $\angle \mathrm{PRQ}=\angle \mathrm{TRS}=\angle \mathrm{R}$

(Common)

So, $\triangle \mathrm{RPQ} \sim \Delta \mathrm{RTS}$.

(AA similarity)

Hence proved

Ans: 

We know that the corresponding portions of two triangles that are congruent to each other are equal.

The two triangles are comparable if one of their angles is equal to one of the other triangle's angles, and the sides that include these angles are proportionate.

For two triangles, this is known as the SAS (Side - Angle - Side) similarity criteria.

In $\triangle \mathrm{ABE}$ and $\triangle \mathrm{ACD}$

$\mathrm{AD}=\mathrm{AE}(\triangle \mathrm{ABE} \cong \Delta \mathrm{ACD} \text { given }) \ldots \ldots \ldots \text { (1) }$

$\mathrm{AB}=\mathrm{AC}(\triangle \mathrm{ABE} \cong \triangle \mathrm{ACD} \text { given })$

Now Consider $\triangle A D E$ and $\triangle A B C$

and $\angle$ DAE $=\angle B A C$ (Common angle)

Thus, $\triangle$ ADE $\sim A$ ABC (SAS criterion)

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.

In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{CDP}$

$[\because \mathrm{CE} \perp \mathrm{AB}$ and $\mathrm{AD} \perp \mathrm{BC} ;$ altitudes $]$

$\angle A P E=\angle C P D$ (Vertically opposite angles)

$\Rightarrow \triangle$ AEP $\sim \triangle$ CPD (AA criterion)

1. $\triangle \mathrm{ABD} \sim \Delta \mathrm{CBE}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.

In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{CBE}$

$\angle \mathrm{ADB}=\angle C E B=90^{\circ}$

$\angle \mathrm{ABD}=\angle C B E \text { (Common angle) }$

$\Rightarrow \triangle \mathrm{ABD} \sim \Delta C B E \text { (AA criterion) }$

1. $\triangle \mathrm{AEP} \sim \triangle \mathrm{ADB}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.

In $\triangle \mathrm{AEP}$ and $\triangle \mathrm{ADB}$

$\angle \mathrm{AEP}=\angle \mathrm{ADB}=9 \mathrm{O}^{\circ}$

$\angle \mathrm{PAE}=\angle \mathrm{BAD} \text { (Common angle) }$

$\Rightarrow \triangle \mathrm{AEP} \sim \triangle \mathrm{ADB} \text { (AA criterion) }$

1. $\Delta \mathrm{PDC} \sim \Delta \mathrm{BEC}$

Ans: When two angles from one triangle are equivalent to two angles from another triangle, the two triangles are said to be comparable.

For two triangles, this is known as the AA similarity criteria.

In $\triangle \mathrm{PDC}$ and $\triangle \mathrm{BEC}$

$\angle \mathrm{PDC}=\angle \mathrm{BEC}=9 \mathrm{O}^{\circ}$

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

$\Rightarrow \triangle \text { PDC } \sim \triangle \mathrm{BEC} \text { (AA criterion)}$

Ans:

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

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

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

$\therefore \Delta \mathrm{ABE} \sim \Delta \mathrm{CFB}$ (By AA similarity criterion)

Ans: In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{AMP}$

$\angle \mathrm{ABC}=\angle \mathrm{AMP}\left(\operatorname{Each} 90^{\circ}\right)$ $\angle \mathrm{A}=\angle \mathrm{A}(\mathrm{Common})$

$\therefore \Delta \mathrm{ABC} \sim \Delta \mathrm{AMP}$ (By AA similarity criterion)

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

Ans: $\text { It is given that } \triangle \mathrm{ABC} \sim \Delta \mathrm{FEG}.$

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

$\text { Since, } \angle \mathrm{ACB}=\angle \mathrm{FGE} $

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

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

$\text { In } \triangle \mathrm{ACD} \text { and } \Delta \mathrm{FGH} \text {, }$

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

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

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

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

Ans: $\text { In } \triangle \mathrm{DCB} \text { and } \triangle \mathrm{HGE} \text {, } $

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

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

$\therefore \triangle \mathrm{DCB} \sim \triangle \mathrm{HGE} \text { (By AA similarity criterion) }$

Ans: In $\Delta \mathrm{DCA}$ and $\Delta \mathrm{HGF}$,

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

$\angle A=\angle F$ (Proved above)

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