Question

Question 13
D is a point on the side BC of a triangle ABC such that $\angle ADC=\angle BAC$. Show that $C{A}^2=CB.CD$.

Answer 1

In $\Delta ADCand\Delta BAC$,
$\angle ADC=\angle BAC\left(Given\right)$
$\angle ACD=\angle BCA$ (Common angle)
$\therefore \Delta ADC\sim \Delta BAC$ (By AAA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
$\therefore \frac{CA}{CB}=\frac{CD}{CA}$
$\Rightarrow C{A}^2=CB.CD$