4. In Fig. 6.36, and . Show that .

Understand the Question In , since , the sides opposite to these angles are equal (). Substituting into the given ratio allows us to relate the corresponding sides of and . Since both triangles share the common angle ($ 1 $ using the SAS similarity criterion. Step 1 · Relate Sides in In Since sides opposite to equal angles in a triangle are equal Step 2 · Substitute into the Given Ratio Given Substitute from equation Taking reciprocals and rearranging terms Step 3 · Apply SAS Similarity Criterion In and Therefore, by SAS similarity criterion Answer > Common Mistakes Overlooking the Substitution: Trying to prove similarity directly with instead of replacing with using the isosceles triangle property of . Incorrect Angle Inclusion: Not verifying that the common angle is the included angle between the two pairs of proportional sides before applying the SAS criterion.

Triangles | Exercise 6.3

Question 4

In Fig. 6.36, $\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}}$ and $\angle 1 = \angle 2$ . Show that $\Delta \text{PQS} \sim \Delta \text{TQR}$ .

Solution

Understand the Question

In $\Delta \text{PQR}$ , since $\angle 1 = \angle 2$ , the sides opposite to these angles are equal ( $\text{PQ} = \text{PR}$ ).
Substituting $\text{PR} = \text{PQ}$ into the given ratio $\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}}$ allows us to relate the corresponding sides of $\Delta \text{PQS}$ and $\Delta \text{TQR}$ .
Since both triangles share the common angle $\angle \text{Q}$ ( $\angle 1$ ) included between these proportional sides, we can prove $\Delta \text{PQS} \sim \Delta \text{TQR}$ using the SAS similarity criterion.

Step 1 · Relate Sides in $\Delta \text{PQR}$

In $\Delta \text{PQR}$ $\angle 1 = \angle 2 \implies \angle \text{PQR} = \angle \text{PRQ}$

Since sides opposite to equal angles in a triangle are equal $\text{PQ} = \text{PR} \quad \dots (i)$

Step 2 · Substitute into the Given Ratio

Diagram 1

Given $\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}}$

Substitute $\text{PR} = \text{PQ}$ from equation $(i)$ $\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PQ}}$

Taking reciprocals and rearranging terms $\dfrac{\text{PQ}}{\text{QT}} = \dfrac{\text{QS}}{\text{QR}} \quad \dots (ii)$

Step 3 · Apply SAS Similarity Criterion

In $\Delta \text{PQS}$ and $\Delta \text{TQR}$ $\dfrac{\text{PQ}}{\text{QT}} = \dfrac{\text{QS}}{\text{QR}} \quad [\text{From } (ii)]$

$\angle \text{PQS} = \angle \text{TQR} = \angle 1 \quad (\text{Common angle})$

Therefore, by SAS similarity criterion $\Delta \text{PQS} \sim \Delta \text{TQR}$

Answer

$\Delta \text{PQS} \sim \Delta \text{TQR}$

Common Mistakes

Overlooking the Substitution: Trying to prove similarity directly with $\text{PR}$ instead of replacing $\text{PR}$ with $\text{PQ}$ using the isosceles triangle property of $\Delta \text{PQR}$ .
Incorrect Angle Inclusion: Not verifying that the common angle $\angle 1$ is the included angle between the two pairs of proportional sides before applying the SAS criterion.

More questions in Exercise 6.3

State which pairs of triangles in Fig. 6.34 are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form :

In Fig. 6.35, $\Delta \text{ODC} \sim \Delta \text{OBA}$ , $\angle \text{BOC} = 125^\circ$ and $\angle \text{CDO} = 70^\circ$ . Find $\angle \text{DOC}$ , $\angle \text{DCO}$ and $\angle \text{OAB}$ .

Diagonals $\text{AC}$ and $\text{BD}$ of a trapezium $\text{ABCD}$ with $\text{AB} \parallel \text{DC}$ intersect each other at the point $\text{O}$ . Using a similarity criterion for two triangles, show that $\dfrac{\text{OA}}{\text{OC}} = \dfrac{\text{OB}}{\text{OD}}$ .

In Fig. 6.36, $\dfrac{\text{QR}}{\text{QS}} = \dfrac{\text{QT}}{\text{PR}}$ and $\angle 1 = \angle 2$ . Show that $\Delta \text{PQS} \sim \Delta \text{TQR}$ .

S and T are points on sides $PR$ and $QR$ of $\Delta \text{PQR}$ such that $\angle \text{P} = \angle \text{RTS}$ . Show that $\Delta \text{RPQ} \sim \Delta \text{RTS}$ .

In Fig. 6.37, if $\Delta \text{ABE} \cong \Delta \text{ACD}$ , show that $\Delta \text{ADE} \sim \Delta \text{ABC}$ .

In Fig. 6.38, altitudes $AD$ and $CE$ of $\Delta \text{ABC}$ intersect each other at the point $P$ . Show that:

(i) $\Delta \text{AEP} \sim \Delta \text{CDP}$

(ii) $\Delta \text{ABD} \sim \Delta \text{CBE}$

(iii) $\Delta \text{AEP} \sim \Delta \text{ADB}$

(iv) $\Delta \text{PDC} \sim \Delta \text{BEC}$

E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that $\Delta \text{ABE} \sim \Delta \text{CFB}$ .

In Fig. 6.39, ABC and AMP are two right triangles, right angled at B and M respectively. Prove that:

(i) $\Delta \text{ABC} \sim \Delta \text{AMP}$

(ii) $\dfrac{\text{CA}}{\text{PA}} = \dfrac{\text{BC}}{\text{MP}}$

Q10

CD and GH are respectively the bisectors of $\angle \text{ACB}$ and $\angle \text{EGF}$ such that D and H lie on sides AB and FE of $\Delta \text{ABC}$ and $\Delta \text{EFG}$ respectively. If $\Delta \text{ABC} \sim \Delta \text{FEG}$ , show that:

(i) $\dfrac{\text{CD}}{\text{GH}} = \dfrac{\text{AC}}{\text{FG}}$

(ii) $\Delta \text{DCB} \sim \Delta \text{HGE}$

(iii) $\Delta \text{DCA} \sim \Delta \text{HGF}$

Q11

In Fig. 6.40, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$ . If $AD \perp BC$ and $EF \perp AC$ , prove that $\Delta ABD \sim \Delta ECF$ .

Q12

Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of $\Delta PQR$ (see Fig. 6.41). Show that $\Delta ABC \sim \Delta PQR$ .

Q13

D is a point on the side $BC$ of a triangle $ABC$ such that $\angle \text{ADC} = \angle \text{BAC}$ . Show that $\text{CA}^2 = \text{CB} \cdot \text{CD}$ .

Q14

Sides $AB$ and $AC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $PR$ and median $PM$ of another triangle $PQR$ . Show that $\Delta ABC \sim \Delta PQR$ .

Q15

A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower.

Q16

If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$ , respectively where $\Delta ABC \sim \Delta PQR$ , prove that

$\dfrac{\text{AB}}{\text{PQ}} = \dfrac{\text{AD}}{\text{PM}}$