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:
\(\triangle CAB\) is a right triangle, right angled at \(A\)
\(AD \perp BC\)
\(BC = 10 \, \text{cm}\), \(CD = 2 \, \text{cm}\)

To prove:
\[ \triangle ADB \sim \triangle CDA \] And find the length of \(AD\).

Proof of similarity:
- In \(\triangle ADB\) and \(\triangle CDA\), both have a right angle:
\[ \angle ADB = 90^\circ, \quad \angle CDA = 90^\circ \]
- They share \(\angle A\).
- Therefore, by AA (Angle-Angle) similarity criterion,
\[ \triangle ADB \sim \triangle CDA \]

Using similarity:
\[ \frac{AD}{CD} = \frac{DB}{AD} \] Cross-multiplied:
\[ AD^2 = CD \times DB \]

Find \(DB\):
\[ DB = BC - CD = 10 - 2 = 8 \, \text{cm} \]

Calculate \(AD\):
\[ AD^2 = 2 \times 8 = 16 \Rightarrow AD = \sqrt{16} = 4 \, \text{cm} \]

Final answer:
Length of \(AD = 4 \, \text{cm}\).