In $\Delta ABC$, $AD \perp BC$ such that $AD^2 = BD \cdot CD$. Prove that $\Delta ABC$ is a right-angled triangle.
Step-by-Step Solution
Key Concept: In $\Delta ABD$ and $\Delta CAD$, $\dfrac{AD}{CD} = \dfrac{BD}{AD}$ and $\angle ADB = \angle ADC = 90^\circ \Rightarrow \Delta ABD \sim \Delta CAD \Rightarrow \angle BAD = \angle ACD$. Sum of angles gives $\angle A = 90^\circ$.
$AD^2 = BD \cdot CD \Rightarrow \dfrac{AD}{CD} = \dfrac{BD}{AD}$. Also $\angle ADB = \angle ADC = 90^\circ$. [1.0 Mark]
By SAS, $\Delta ABD \sim \Delta CAD \Rightarrow \angle BAD = \angle C$ and \angle B = \angle CAD$. [0.5 Mark]
$\angle A = \angle BAD + \angle CAD = \angle C + \angle B$. In $\Delta ABC$, $\angle A + \angle B + \angle C = 180^\circ \Rightarrow 2\angle A = 180^\circ \Rightarrow \angle A = 90^\circ$. Proved! [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Proving $\Delta ABD \sim \Delta CAD$: 1.0 Mark
Showing $\angle A = \angle B + \angle C$: 0.5 Mark
Concluding $\angle A = 90^\circ$: 0.5 Mark
Correct Answer: