PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM² = QM×MR. (CBSE 2004, 06, 09)

To Prove: PM² = QM x MR
Given: PM is perpendicular to QR and PQR is a right angled triangle at P
In triangles QMP and PMR,
PM = common
And QM = MR
Now, by RHS rule Both the triangle are congruent 
Therefore,
∠MPR = ∠MPQ
Let, ∠MPR = ∠MPQ =  x  ..... (1)

In triangle MPR,
As sum of triangles is 180^o ,
∠MRP + ∠MPR + ∠RMP = 180^o

⇒ ∠MRP = 180^o – 90^o – x

 ⇒∠MRP = 90^o – x  .... (2)

Similarly,

In triangle MPQ,
∠MPQ + ∠PMQ + ∠MQP = 180^o
∠MQP = 180^o  - ∠PMQ - ∠MPQ
∠MQP = 180^o  - 90^o - ∠MPQ
∠MQP = 90^o - ∠MPR

∠MQP = 90^o – x  .... (3)

In triangle QMP and PMR,

∠MQP = ∠MRP  (from 2 and 3)

∠PMQ = ∠RMP  (each 90^{o )}

∠MPQ = ∠MPR  (from 1)

Therefore,

ΔQMP ~ΔPMR (By Angle-Angle-Angle similarity)

PM² = QM×MR
Hence, Proved.