Question

In given figure ABC is a triangle in which AB=AC and D is a point on AC such that ${BC}^2=AC\times CD$. Prove that BD=BC.

Answer 1

Given$△ABC$ in which $AB=AC$ and D is a point on the side AC such that

${BC}^2=AC\times CD$

To prove: $BD=BC$

construction: $joinBD$

we have,

${BC}^2=AC\times CD$

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

Thus, in $△ABC$ and $△BDC$, we have

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

$\angle C=\angle C$ [Common]

$\therefore$ $△ABC\sim △BDC$ [By SAS criterion of similarity]

$\Rightarrow$ $\frac{AB}{BD}=\frac{BC}{DC}$

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

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

From (i) and (ii), we get

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

$\Rightarrow BD=BC$ $\left[Henceproved\right]$