Question

2. PQR is a triangle right angled at P and M is a
point on QR such that PM IQR. Show that
PM = QM. MR.
3
​

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≈)

BD²=AD×CD