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.
Generated by claude-sonnet-4-6 · 2026-06-15 10:27 · grounding rag
Model Answer
Proof that △ADB ~ △CDA:
In △ADB and △CDA:
∠ADB = ∠CDA = 90° (AD ⊥ BC)
∠ABD = ∠CAD (In △ABC, ∠A = 90°, so ∠B + ∠C = 90°; also in △ADB, ∠ADB = 90°, so ∠ABD + ∠DAB = 90°; and in △CAD, ∠CDA = 90°, so ∠ACD + ∠DAC = 90°. Since ∠ABD = ∠ACD = ∠B is common to both... )
More directly:
∠DBA = ∠DAC ... (each equals 90° − ∠C, since in △ADB: ∠B + ∠DAB = 90°, and in △CAB: ∠B + ∠C = 90°, so ∠DAB = ∠C, i.e. ∠DBA = 90° − ∠C and ∠DAC = 90° − ∠C)
∠ADB = ∠CDA = 90° (given)
∴ By AA similarity criterion, △ADB ~ △CDA (proved)
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Finding AD, given BC = 10 cm, CD = 2 cm:
DB = BC − CD = 10 − 2 = 8 cm
Since △ADB ~ △CDA, corresponding sides are proportional:
$$\frac{AD}{CD} = \frac{DB}{AD}$$
$$AD^2 = CD \times DB = 2 \times 8 = 16$$
$$AD = 4 \text{ cm}$$
Source: Triangles, Section 6.4 (AA Similarity Criterion)
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Explanation
- Two angles to prove AA similarity: Always identify the two pairs of equal angles clearly. Here, the right angles at D are obvious; the second pair requires noting that in a right triangle, the acute angles are complementary, so ∠B = 90° − ∠C, and ∠DAC = 90° − ∠C as well (since ∠DAB + ∠C = 90° from △CAB).
- Order of vertices matters: Writing △ADB ~ △CDA means A↔C, D↔D, B↔A — set up the ratio accordingly as $\frac{AD}{CD} = \frac{DB}{DA}$, giving $AD^2 = CD × DB$.
- Examiners look for: Correct angle identification with reasons, correct similarity statement with proper vertex correspondence, and clean use of the proportionality to find AD.