# In a triangle ABC, AB=AC and D is a point on side AC such that BCxBC=ACxCD. How to prove that BD=BC?

Aug 24, 2016

Given

$\text{In } \Delta A B C$

$A B = A C \mathmr{and} D \text{ is a point on"AC " such that}$

$B C \times B C = A C \times A D$

$\text{We are to prove } B D = B C$

Proof

Rearrenging the given relation

$B C \times B C = A C \times A D \text{ }$ We can write

$\frac{B C}{C D} = \frac{A C}{B C} \to \Delta A B C \text{ similar } \Delta B D C$

Their corresponding angle pairs are:

$1. \angle B A C \text{= corresponding } \angle D B C$

$2. \angle A B C \text{= corresponding } \angle B D C$

$3. \angle A C B \text{ =corresponding } \angle D C B$

So as per above relation 2 we have
$\angle A B C = \text{ corresponding } \angle B D C$

$\text{Again in} \Delta A B C$

$A B = A C \to \angle A B C = \angle A C B = \angle D C B$

$\therefore \text{In } \Delta B D C , \angle B D C = \angle B C D$

$\to B D = B C$

Alternative way

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

$\frac{B C}{C D} = \frac{A C}{B C} = \frac{A B}{B D}$

From this relation we have

$\frac{A C}{B C} = \frac{A B}{B D}$

$\implies \frac{A C}{B C} = \frac{A C}{B D} \to \text{As "AB=AC" given}$

$\implies \frac{1}{B C} = \frac{1}{B D}$

$\implies B C = B D$

Proved

Hope, this will help