Question

In triangle abc ab=ac and d is a mid point on side ac,such that bc square =ac ×cd. Prove that bd =bc

Answer 1

Solution;

Given that,

In ΔABC,
AB = AC
And D is a point on AC such that BC² = AC×AD

To prove, BD=BC
Proofs:

Rearranging the given relation:

BC×BC = AC×AD  We can write as:

BC/CD = AC/BC → ΔABC similar ΔBDC

Their corresponding angle pairs are:

1.∠BAC = ∠DBC [Being Corresponding angle]

2.∠ABC =∠BDC [Being Corresponding angle]

3.∠ACB = ∠DCB [Being Corresponding angle]

So as per above relation 2 we have:

∠ABC =∠BDC

Again in ΔABC

AB=AC→∠ABC=∠ACB=∠DCB

∴In ΔBDC, ∠BDC = ∠BCD

⇒ BD = BC