In $\Delta ABC$, $\angle A = 90^\circ$ and $AD \perp BC$. Prove that $AD^2 = BD \cdot CD$.
Step-by-Step Solution
Key Concept: $\Delta ABD \sim \Delta CAD$ by AA $\Rightarrow \dfrac{AD}{CD} = \dfrac{BD}{AD} \Rightarrow AD^2 = BD \cdot CD$.
In $\Delta ABD$ and $\Delta CAD$, $\angle BAD = \angle C$ and $\angle ADB = \angle ADC = 90^\circ \Rightarrow \Delta ABD \sim \Delta CAD$. [1.0 Mark]
$\dfrac{AD}{CD} = \dfrac{BD}{AD} \Rightarrow AD^2 = BD \cdot CD$. Proved! [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Proving $\Delta ABD \sim \Delta CAD$: 1.0 Mark
Equating ratios to get $AD^2 = BD \cdot CD$: 1.0 Mark
Correct Answer: