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

We will use the following similarity criterion: - SAS (Side-Angle-Side): If two sides of one triangle are proportional to two corresponding sides of another triangle, and the included angles are equal, the triangles are similar. We will use the given information to prove the similarity of the two triangles. Step 1 — Identify equal sides We are given that . From the diagram, is . Also, is . So, in , we have . Sides opposite to equal angles are equal. Therefore, . Step 2 — Substitute into the given ratio We are given the ratio . From Step 1, we know that . Let's substitute for in the given ratio. Now, let's rearrange this ratio to match the SAS criterion. Step 3 — Apply SAS similarity criterion Consider and . We have . This is because is common to both triangles. From Step 2, we found the ratio of sides: Since one angle is equal and the sides including this angle are proportional, the triangles are similar by SAS criterion. Answer >

Triangles | Exercise 6.3

Question 4

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

Solution

We will use the following similarity criterion:

SAS (Side-Angle-Side): If two sides of one triangle are proportional to two corresponding sides of another triangle, and the included angles are equal, the triangles are similar.

We will use the given information to prove the similarity of the two triangles.

Step 1 — Identify equal sides

We are given that $\angle 1 = \angle 2$ . From the diagram, $\angle 1$ is $\angle \text{PQR}$ . Also, $\angle 2$ is $\angle \text{PRQ}$ . So, in $\Delta \text{PQR}$ , we have $\angle \text{PQR} = \angle \text{PRQ}$ . Sides opposite to equal angles are equal. Therefore, $\text{PQ} = \text{PR}$ .

Step 2 — Substitute into the given ratio

We are given the ratio $\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}}$ . From Step 1, we know that $\text{PR} = \text{PQ}$ . Let's substitute $\text{PQ}$ for $\text{PR}$ in the given ratio.

$\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PQ}}$

Now, let's rearrange this ratio to match the SAS criterion.

$\frac{\text{PQ}}{\text{QT}} = \frac{\text{QS}}{\text{QR}}$

Diagram 1

Step 3 — Apply SAS similarity criterion

Consider $\Delta \text{PQS}$ and $\Delta \text{TQR}$ . We have $\angle \text{PQS} = \angle \text{TQR}$ . This is because $\angle 1$ is common to both triangles. From Step 2, we found the ratio of sides:

$\frac{\text{PQ}}{\text{QT}} = \frac{\text{QS}}{\text{QR}}$

Since one angle is equal and the sides including this angle are proportional, the triangles are similar by SAS criterion.

Answer

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

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

In Fig. 6.36, $\frac{\text{QR}}{\text{QS}} = \frac{\text{QT}}{\text{PR}}$ 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) $\frac{\text{CA}}{\text{PA}} = \frac{\text{BC}}{\text{MP}}$

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) $\frac{\text{CD}}{\text{GH}} = \frac{\text{AC}}{\text{FG}}$ (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 $\angle$ ADC = $\angle$ BAC. Show that $\text{CA}^2 = \text{CB.CD}$ .

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 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 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

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