Question

In given figure 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 △ABC in which AB = AC and D is a point on the side AC such that

BC² = AC × CD

To prove: BD = BC

construction: join BD

we have,

BC² = AC × CD

⇒ \(\frac{BC}{CD} = \frac{AC}{BC}\)   ......(i)

Thus, in △ABC and △BDC, we have

\( \frac{AC}{BC}=\frac{BC}{CD} \)    [From (i)]  and,

∠C = ∠C      [Common]

∴ △ABC ∼ △BDC    [By SAS criterion of similarity]

⇒ \(\frac{AB}{BD} = \frac{BC}{DC}\)

⇒ \( \frac{AC}{BD}=\frac{BC}{CD} \)   [∵ AB = AC]

⇒ \( \frac{AC}{BC}=\frac{BD}{CD} \)   ......(ii)

From (i) and (ii), we get

\( \frac{BC}{CD}=\frac{BD}{CD} \)

⇒ BD = BC   [Hence proved]