Question

In a triangle $ABC,AB=AC$ and $D$ is a point on side $AC$ such that $B{C}^2=AC\times CD$ prove that $BD=BC$

Answer 1

As $B{C}^2=AC\times CD$, then,

$\frac{BC}{AC}=\frac{CD}{BC}$

In $\Delta ABC$ and $\Delta BDC$,

$\frac{BC}{AC}=\frac{CD}{BC}$

And,

$\angle C=\angle C$

Therefore, by SAS property,

$\Delta ABC\sim \Delta BDC$

So,

$\frac{AB}{AC}=\frac{BD}{BC}$ [C.S.C.T.]

$\frac{AB}{AB}=\frac{BD}{BC}$ [$\because AB=AC$]

$1=\frac{BD}{BC}$

$BD=BC$ $\left[henceproved\right]$