In ΔABC, D is a point on side BC such that ∠ADC = ∠BAC. If CA = 12 cm, CD = 8 cm, then CB (in cm) = ?

In ΔABC,
∠BAC = ∠ADC, (given)
In, ΔABC & ΔADC,
∠BAC = ∠ADC (given)
∠C = ∠C (common)
$$\eqalign{ & \therefore \Delta {\text{ABC}} \sim \Delta {\text{DAC}}\,\,\,\left( {{\text{A}} - {\text{A}}\,{\text{theorem}}} \right) \cr & \frac{{{\text{BC}}}}{{{\text{AC}}}} = \frac{{{\text{AC}}}}{{{\text{DC}}}} \cr & \Rightarrow \frac{{{\text{BC}}}}{{12}} = \frac{{12}}{8} \cr & \Rightarrow {\text{BC}} = \frac{{144}}{8} = 18{\text{ cm}} \cr} $$