In $\Delta ABC$, $\angle A = 90^\circ$ and $AD \perp BC$. Prove that $AD^2 = BD \cdot CD$.
Step-by-Step Solution
Key Concept: Prove $\Delta ABD \sim \Delta CAD$ by angle matching.
Stepwise Solution:
In right $\Delta ABD$: $\angle DAB + \angle B = 90^\circ$. Also $\angle DAB + \angle CAD = 90^\circ \Rightarrow \angle B = \angle CAD$. [0.5 Mark]
In $\Delta ABD$ and $\Delta CAD$:
$\angle ADB = \angle ADC = 90^\circ$
$\angle B = \angle CAD$. [0.5 Mark]
By AA similarity, $\Delta ABD \sim \Delta CAD$. [0.5 Mark]
Therefore $\dfrac{AD}{CD} = \dfrac{BD}{AD} \Rightarrow AD^2 = BD \cdot CD$. Proved! [0.5 Mark]
Marking Scheme:
• Proving angle equality $\angle B = \angle CAD$: 0.5 Mark
• Establishing similarity $\Delta ABD \sim \Delta CAD$: 0.5 Mark
• Writing corresponding side ratios: 0.5 Mark
• Concluding $AD^2 = BD \cdot CD$: 0.5 Mark
Correct Answer: