In $\Delta ABC$, $D$ is a point on $BC$ such that $\angle ADC = \angle BAC$. Then $CA^2$ is equal to:
(a) $CB \cdot CD$
(b) $CB \cdot BD$
(c) $CD \cdot BD$
(d) $AB \cdot AC$
Step-by-Step Solution
Key Concept: $\Delta ABC \sim \Delta DAC \Rightarrow \dfrac{CA}{CD} = \dfrac{CB}{CA} \Rightarrow CA^2 = CB \cdot CD$.
$\Delta ABC \sim \Delta DAC$ by AA similarity. $\dfrac{CA}{CD} = \dfrac{CB}{CA} \Rightarrow CA^2 = CB \cdot CD$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Proving $\Delta ABC \sim \Delta DAC \Rightarrow CA^2 = CB \cdot CD$: 1.0 Mark
Correct Answer: $CB \cdot CD$