Question

In ΔABC and ΔPQR, AD and PS are altitudes such that ΔABD ~ ΔPQS and ΔACD ~ ΔPRS. Prove that ΔABC ~ ΔPQR.

Answer 1

Given: ΔABC and ΔPQR with AD and PS as altitudes (D on BC, S on QR). ΔABD ~ ΔPQS and ΔACD ~ ΔPRS.

To Prove: ΔABC ~ ΔPQR.

Proof (Step-wise):

1. Since AD and PS are altitudes, AD ⟂ BC and PS ⟂ QR.

Hence ∠ADB = 90° and ∠PSQ = 90°.

2. From ΔABD ~ ΔPQS (given) the correspondence is A ↔ P, B ↔ Q, D ↔ S.

Thus corresponding angles are equal:

(i) ∠ABD = ∠PQS and (ii) ∠BAD = ∠QPS.

Note: ∠ABD is the same as ∠ABC because D lies on BC; similarly ∠PQS is the same as ∠PQR because S lies on QR. Therefore ∠ABC = ∠PQR.

3. From ΔACD ~ ΔPRS (given) the correspondence is A ↔ P, C ↔ R, D ↔ S.

Thus corresponding angles are equal:

(i) ∠CAD = ∠RPS and (ii) ∠ACD = ∠PRS.

Note: ∠CAD is the same as ∠BAC (angles at A split by AD) and ∠RPS is the same as ∠QPR (angles at P split by PS).

Therefore ∠BAC = ∠QPR.

4. From steps 2 and 3 we have two pairs of corresponding angles equal:

∠ABC = ∠PQR and ∠BAC = ∠QPR

5. By the AA (angle–angle) similarity criterion, if two angles of one triangle are equal to two angles of another, the triangles are similar.

Therefore ΔABC ~ ΔPQR.