6. In Fig. 6.37, if , 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 properties of congruent triangles to find equal sides. Then we will use these equal sides to show the ratio of corresponding sides is equal. Finally, we will use the SAS similarity criterion. Step 1 — Identify equal parts We are given that is congruent to . Congruent triangles have corresponding parts equal. So, corresponding sides are equal. Step 2 — Form ratios of sides We have . We also have . Let's divide the first equality by the second one. > Step 3 — Apply SAS similarity Consider and . Both triangles share the common angle . We have already shown that the ratio of the sides including this angle is equal. Therefore, by the SAS similarity criterion, the triangles are similar. Answer > (i) is shown by SAS similarity.

Triangles | Exercise 6.3

Question 6

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

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 properties of congruent triangles to find equal sides. Then we will use these equal sides to show the ratio of corresponding sides is equal. Finally, we will use the SAS similarity criterion.

Step 1 — Identify equal parts

We are given that $\Delta \text{ABE}$ is congruent to $\Delta \text{ACD}$ . Congruent triangles have corresponding parts equal. So, corresponding sides are equal.

$\text{AB} = \text{AC} \quad (\text{CPCTC})$

$\text{AD} = \text{AE} \quad (\text{CPCTC})$

Diagram 1

Step 2 — Form ratios of sides

We have $\text{AD} = \text{AE}$ . We also have $\text{AB} = \text{AC}$ . Let's divide the first equality by the second one.

$\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}}$

$\boxed{\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}}}$

Step 3 — Apply SAS similarity

Consider $\Delta \text{ADE}$ and $\Delta \text{ABC}$ . Both triangles share the common angle $\angle \text{A}$ . We have already shown that the ratio of the sides including this angle is equal.

$\angle \text{DAE} = \angle \text{BAC} \quad (\text{Common angle})$

$\frac{\text{AD}}{\text{AB}} = \frac{\text{AE}}{\text{AC}} \quad (\text{From Step 2})$

Therefore, by the SAS similarity criterion, the triangles are similar.

$\Delta \text{ADE} \sim \Delta \text{ABC}$

Answer

(i) $\Delta \text{ADE} \sim \Delta \text{ABC}$ is shown by SAS similarity.

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

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