Question

\(D\) is a point on the side \(BC\) of \(\triangle ABC\) such that \(\angle ADC = \angle BAC\). Show that:
\(AC^2 = BC \times DC\)

Answer 1

We are given that \(\angle ADC = \angle BAC\), which means triangle \(ADC\) is similar to triangle \(ABC\) by the AA (Angle-Angle) similarity criterion.

We can apply the following proportionality rule:

\[ \frac{AC}{BC} = \frac{DC}{AC} \]

Now, cross-multiply to get:

\[ AC^2 = BC \times DC \]

This is the required result.