Question

if in an isoceles triangle ABC,AB=AC such that Bc square = AC×CD then prove BD=BC​

Answer 1

Answer:

BD=BC​

Step-by-step explanation:

Given in 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

Answer 2

Hello once again

Given: A △ABC in which AB = AC. D is a point on AC such that BC2 = AC × CD.

To prove : BD = BC  

Proof : Since BC2 = AC × CD

Therefore BC × BC = AC × CD

AC/BC = BC/CD .......(i)

Also ∠ACB = ∠BCD

Since △ABC ~ △BDC [By SAS Axiom of similar triangles]

AB/AC = BD/BC ........(ii)

But AB = AC (Given) .........(iii)

From (i),(ii) and (iii) we get  

BD = BC.

Hope hels

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