S and T are points on sides and of such that . Show that .

Understand the Question According to the AA (Angle-Angle) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar. In and : (or ) is given to be equal to . Both triangles share as a common angle. By establishing these two pairs of equal angles, we can prove that . Step 1 · Prove Similarity Using AA Criterion Consider and . In and : Therefore, by the AA similarity criterion: Answer > Common Mistakes Incorrect Vertex Correspondence: Writing instead of correctly matching equal angles: , , and , which gives . Assuming Parallel Lines: Incorrectly assuming . The angle given is , which means the sides are in an inverted correspondence, not parallel.

Triangles | Exercise 6.3

Question 5

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

Solution

Understand the Question

According to the AA (Angle-Angle) similarity criterion, if two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
In $\Delta \text{RPQ}$ $Δ RPQ$ and $\Delta \text{RTS}$ $Δ RTS$ :
- $\angle \text{P}$ (or $\angle \text{QPR}$ ) is given to be equal to $\angle \text{RTS}$ .
- Both triangles share $\angle \text{R}$ as a common angle.
By establishing these two pairs of equal angles, we can prove that $\Delta \text{RPQ} \sim \Delta \text{RTS}$ .

Step 1 · Prove Similarity Using AA Criterion

Consider $\Delta \text{RPQ}$ and $\Delta \text{RTS}$ . Diagram 1

In $\Delta \text{RPQ}$ and $\Delta \text{RTS}$ : $\angle \text{QPR} = \angle \text{RTS} \quad (\text{Given})$ $\angle \text{PRQ} = \angle \text{TRS} \quad (\text{Common angle})$

Therefore, by the AA similarity criterion: $\Delta \text{RPQ} \sim \Delta \text{RTS}$

Answer

$\Delta \text{RPQ} \sim \Delta \text{RTS}$

Common Mistakes

Incorrect Vertex Correspondence: Writing $\Delta \text{PQR} \sim \Delta \text{STR}$ instead of correctly matching equal angles: $\text{R} \leftrightarrow \text{R}$ , $\text{P} \leftrightarrow \text{T}$ , and $\text{Q} \leftrightarrow \text{S}$ , which gives $\Delta \text{RPQ} \sim \Delta \text{RTS}$ .
Assuming Parallel Lines: Incorrectly assuming $ST \parallel PQ$ . The angle given is $\angle \text{RTS} = \angle \text{P}$ , which means the sides are in an inverted correspondence, not parallel.

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

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