Question

In the following figure, PQR is a triangle in which PS and QT are altitudes from P and Q respectively, intersecting each other at M. Prove that ΔQSM ~ ΔРТМ.

Answer 1

Given:

In triangle PQR, PS ⟂ QR and QT ⟂ PR.

PS and QT meet at M (so M is the intersection of the two altitudes).

To Prove: ΔQSM ~ ΔPTM.

Proof [Step-wise]:

1. Since S is the foot of the altitude from P on QR, PS ⟂ QR.

Therefore SM (a part of PS) is perpendicular to QS (a part of QR).

Hence ∠QSM = 90°.

2. Since T is the foot of the altitude from Q on PR, QT ⟂ PR.

Therefore TM (a part of QT) is perpendicular to PT (a part of PR).

Hence ∠PTM = 90°.

3. From (1) and (2), we get

∠QSM = ∠PTM   ...(Both right angles)

4. At M, MQ lies along QT and MS lies along PS, so ∠QMS is the angle between QT and PS. Also at M, MP lies along PS and MT lies along QT, so ∠PMT is the angle between PS and QT.

Thus ∠QMS = ∠PMT.

5. We have two pairs of equal angles: ∠QSM = ∠PTM and ∠QMS = ∠PMT. 

Therefore by the AA (angle–angle) similarity criterion, ΔQSM ~ ΔPTM.

ΔQSM is similar to ΔPTM (correspondence Q ↔ P, S ↔ T, M ↔ M).