Question

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Answer 1

Answer:

Let's construct a diagram according to the given question.

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM.MR.

We know that if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to each other and to the whole triangle.

In ΔPQR we have,

∠QPR = 90° and PM ⊥ QR

In ΔPQR and ΔMQP

∠QPR = ∠QMP = 90°

∠PQR = ∠MQP (commom angle)

⇒ ΔPQR ~ ΔMQP (AA Similarity) --------(1)

In ΔPQR and ΔMPR

∠QPR = ∠PMR = 90°

∠PRQ = ∠PRM (commom angle)

⇒ ΔPQR ~ ΔMPR  (AA Similarity) --------(2)

From equation (1) and (2)

ΔMQP ~ ΔMPR

PM / MR = QM / PM (corresponding sides of similar triangles are proportional)

⇒ PM2 = QM.MR

Hence proved.

Step-by-step explanation: