Question

In the adjoining figure, two sides AB, AC and altitude AM of ΔABC are respectively equal to two sides PQ, PR and altitude PN of ΔPQR. Prove that ΔABC ≅ ΔPQR.

Answer 1

Given: AB = PQ, AC = PR and AM = PN, where AM ⟂ BC and PN ⟂ QR (AM and PN are altitudes).

To Prove: ΔABC ≅ ΔPQR.

Proof [Step-wise]:

1. Since AM and PN are altitudes, ∠AMB = ∠PNQ = 90° and ∠AMC = ∠PNR = 90°.

2. Consider right triangles ΔABM and ΔPQN:

Hypotenuse AB = PQ.   ...(Given)

One leg AM = PN.   ...(Given)

Right angles at M and N.

Hence, ΔABM ≅ ΔPQN by the RHS (right-hypotenuse-side) congruence criterion.

From this congruence,

BM = QN   ...(Corresponding parts)

3. Consider right triangles ΔAMC and ΔPNR:

Hypotenuse AC = PR.   ...(Given)

One leg AM = PN.   ...(Given)

Right angles at M and N.

Hence, ΔAMC ≅ ΔPNR by RHS.

From this congruence,

MC = NR

4. Add the equal segments found:

BC = BM + MC

= QN + NR

= QR

So, BC = QR.

5. Now AB = PQ, BC = QR and AC = PR, so all three corresponding sides of ΔABC and ΔPQR are equal.

Therefore, ΔABC ≅ ΔPQR by SSS congruence.

ΔABC is congruent to ΔPQR.