Question

In ΔPQR, ∠QPR = 90° and PM ⊥ QR. Prove that : PM² = QM.RM.

Answer 1

Given:

In triangle PQR,

∠QPR = 90°,

So, P is the right angle.

PM ⟂ QR M is the foot of the perpendicular from P to the hypotenuse QR.

To Prove:

PM² = QM.RM.

Proof [Step-wise]:

1. Note the three triangles determined by the altitude PM:

ΔPQM, ΔPMR and ΔPQR.

Each pair of these triangles has one right angle and they share an acute angle.

So, by AA they are similar to one another.

2. In particular, ΔPQM is similar to ΔPMR.

From that similarity, the corresponding sides are in proportion.

Choose the correspondence so that PM corresponds to PM.

QM corresponds to RM and QM corresponds to RM. 

The similarity gives the ratio of the leg adjacent to the right angle to the other leg. 

Concretely the similarity yields `(PM)/(QM) = (RM)/(PM)`.

3. Rearranging the proportion in step 2 gives PM² = QM.RM.

Therefore, PM² = QM.RM, as required.