Question

In the adjoining figure, \( \triangle CAB \) is a right triangle, right angled at A and \( AD \perp BC \). Prove that \( \triangle ADB \sim \triangle CDA \). Further, if  \( BC = 10 \text{ cm} \) and \( CD = 2 \text{ cm} \), find the length of } \( AD \).

Hint:

To prove similarity in right-angled triangles, look for AA criterion. Use geometric mean theorem: in right triangle, altitude = \( \sqrt{CD \cdot DB} \).

Answer 1

Given: \( \triangle CAB \) right angled at \( A \), and \( AD \perp BC \) To Prove: \( \triangle ADB \sim \triangle CDA \) Proof: In \( \triangle ADB \) and \( \triangle CDA \): \begin{itemize} \item \( \angle ADB = \angle CDA = 90^\circ \) \item \( \angle DAB = \angle DAC \) (common angle) \end{itemize} \[ \Rightarrow \triangle ADB \sim \triangle CDA \quad \text{(AA similarity)} \] Now, using similarity: \[ \frac{AD}{CD} = \frac{CD}{DB} \Rightarrow AD^2 = CD \cdot DB \] Also, \( BC = BD + CD = 10 \Rightarrow BD = 8 \) \[ AD^2 = 2 \cdot 8 = 16 \Rightarrow AD = \sqrt{16} = \boxed{4 \text{ cm}} \]