Question

$please \: help \: asap$
ABC is a triangle in which AB=AC and D is a point on AC such that BC²=AC×CD.Prove that BD=BC.​

Answer 1

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

Hope, this will help

Answer 2

ΔABC, AB = AC D is a point on AC such that BC2 = AC × AD In ΔABC and ΔBDC ∠C = ∠C (Common angle) ∴ ΔABC ~ ΔBDC [By SAS similarity criterion] [Since triangles are similar, corresponding sides are proportional] From (1) and (2), we get ∴ BC = BD