Question

D is a point on the side BC of $\Delta ABC$ such that $\angle ADC=\angle BAC$. Prove that $\frac{CB}{CA}$ = $\frac{CA}{DC}$ or $C{A}^2$= $CB\times CD$.

[2 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark

In $\Delta ABC$ and $\Delta DAC$, we have

$\angle ADC=\angle BAC$ [Given]

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

$\therefore \Delta ABC\sim \Delta DAC$ [AA-criterion of similarity]

$\Rightarrow \frac{AB}{DA}=\frac{BC}{AC}=\frac{AC}{DC}$

$\Rightarrow \frac{CB}{CA}=\frac{CA}{DC}$

$\Rightarrow C{A}^2=CB\times CD$.