Question

ABC is a triangle in which AB=AC and D is a point on AC such that BC square = AC×CD. prove that BD=BC.


please answer this....

Answer 1

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

To prove : BD=BC

Proof
Rearrenging the given relation, BC×BC=AC×AD

We can write
BC/CD = AC/BC => ΔABC similar ΔBDC

Their corresponding angle pairs are:
∠BAC = ∠DBC
∠ABC = ∠BDC
∠ACB = ∠DCB

So as per above relation 2 we have
∠ABC = ∠BDC

Again in ΔABC, AB=AC =>∠ABC=∠ACB=∠DCB
∴ In ΔBDC, ∠BDC=∠BCD
=> BD = BC