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 $BC = 10$ cm and $CD = 2$ cm, find the length of AD.

Hint:

Use perpendicular height in right triangles and apply geometric mean theorem.

Answer 1

Given:
\( \angle CAB = 90^\circ \), and \( AD \perp BC \)

Prove:
\( \triangle ADB \sim \triangle CDA \)

Proof of similarity:
In triangles \( \triangle ADB \) and \( \triangle CDA \):
- \( \angle ADB = \angle CDA = 90^\circ \) (right angles)
- \( \angle BAD = \angle DAC \) (common)
By AA (Angle-Angle) criterion,
\[ \triangle ADB \sim \triangle CDA \]
Application of similar triangles:
Property of similar triangles:
\[ \text{(Altitude)}^2 = \text{Product of segments of hypotenuse} \] So, \( AD^2 = CD \cdot DB \)

Using provided values:
- \( BC = 10 \, \text{cm} \)
- \( CD = 2 \, \text{cm} \)
- \( BD = BC - CD = 10 - 2 = 8 \, \text{cm} \)

Calculate \( AD \):
\[ AD^2 = CD \cdot DB = 2 \cdot 8 = 16 \Rightarrow AD = \sqrt{16} = 4 \, \text{cm} \]
Answer:
\( AD = \mathbf{4 \, \text{cm}} \)