Question

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

Answer 1

We have,

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

$\Rightarrow$ $BC\times BC=AC\times CD$ and $\angle B=\angle C$

$\Rightarrow$ $\frac{BC}{AC}=\frac{CD}{BC}$ and $\angle B=\angle C$

$\Rightarrow$ $\frac{BC}{CA}=\frac{DC}{CB}$ and $\angle B=\angle C$

So, by SAS-criterion of similarity, we have

$△BCA\sim △DCB$

$\Rightarrow$ $\frac{BC}{DC}=\frac{CA}{CB}=\frac{BA}{DB}$

$\Rightarrow$ $\frac{CA}{CB}=\frac{BA}{DB}$

$\Rightarrow$ $\frac{BA}{CA}=\frac{DB}{CB}$

$\Rightarrow$ $1=\frac{DB}{CB}$ [$\because$ BA=CA]

$\Rightarrow$ $DB=CB$

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