Question

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

Answer 1

Lets take ΔABC B=90 BD perpend. to AC
In ΔABC, ∧B=90°
∧C+∧A=90
let ∧A=x
Therefore, ∧C=90-∧A
⇒90-x
In ΔADB,
∧D=90
∧A+∧ABD=90
∧ABD=90-∧A
⇒90-x

In ΔADB & ΔBDC
∧D=∧D=90
∧ABD=∧C       (=90-x)
ΔADB≈ΔBDC   (AA≈)
$\frac{BD}{CD} = \frac{AD}{BD}$
BD²=AD×CD