Question

In triangle ABC AB = AC and D is a point on side AC such that BC² = AC × CD prove that BD = BC

Answer 1

Answer:

Step-by-step explanation:

Given

In ΔABC

AB=ACandD is a point onAC such that

BC×BC=AC×AD

We are to prove BD=BC

Proof

Rearrenging the given relation

BC×BC=AC×AD We can write

BCCD=ACBC→ΔABC similar ΔBDC

Their corresponding angle pairs are:

1.∠BAC= corresponding ∠DBC

2.∠ABC= corresponding  ∠BDC

3.∠ACB =corresponding  ∠DCB

So as per above relation 2 we have

∠ABC= corresponding ∠BDC

Again inΔABC

AB=AC→∠ABC=∠ACB=∠DCB

∴In ΔBDC,∠BDC=∠BCD

→BD=BC

Alternative way

The ratio of corresponding sides may be written in extended way as follows

BCCD=ACBC=ABBD

From this relation we have

ACBC=ABBD

⇒ACBC=ACBD→As AB=AC given

⇒1BC=1BD

⇒BC=BD

Proved