Question

In the given figure, two triangles ABC and PQR are shown such that ∠A = ∠P and ∠C = ∠R. If AD ⊥ BC and PS ⊥ QR, then prove that (i) ΔADB ~ ΔPSQ (ii) AD × QS = BD × PS.

Answer 1

Given:

Two triangles ABC and PQR such that ∠A = ∠P and ∠C = ∠R.

AD ⊥ BC and PS ⊥ QR

To Prove:

1. ΔADB ~ ΔPSQ 
2. AD × QS = BD × PS

(i) Prove ΔADB ~ ΔPSQ 

Step 1: Identify the right angles

Since AD ⊥ BC, ∠ADB = 90°.

Since PS ⊥ QR, ∠PSQ = 90°.

Step 2: Given angles

∠A = ∠P   ...(Given)

Step 3: Use Angle-Angle (AA) similarity criterion

In ΔADB and ΔPSQ, we have:

∠ADB = ∠PSQ = 90°

∠A = ∠P   ...(Given)

Therefore, by AA similarity criterion,

ΔADB ~ ΔPSQ 

(ii) Prove AD × QS = BD × PS

Since ΔADB ∼ ΔPSQ, corresponding sides are proportional:

`(AD)/(PS) = (BD)/(QS) = (AB)/(PQ)`

From the first two ratios:

`(AD)/(PS) = (BD)/(QS)`

⇒ AD × QS = BD × PS

This proves the required relation.

Hence, both parts (i) and (ii) are proved.