Question

In the adjoining figure, ΔCAB is a right triangle, right angled at A and AD ⊥ BC. Prove that ΔADB ~ ΔCDA. Further, if BC = 10 cm and CD = 2 cm, find the length of AD.

Answer 1

Given:

• ΔCAB is a right-angled triangle with ∠A = 90°.
• AD ⊥ BC, meaning AD is the altitude drawn from A to BC.
• We need to prove that ΔADB ~ ΔCDA (similarity proof).
• Given BC = 10 cm and CD = 2 cm, find AD.

To prove similarity between ΔADB and ΔCDA, we use the AA (Angle-Angle) Similarity Criterion. 

Common Angle: ∠D is common in ΔADB and ΔCDA.

Right Angles: ∠ADB = ∠CDA = 90° (Given AD ⊥ BC).

Since two angles are equal, the triangles are similar by AA similarity.

△ADB ∼ △CDA

Since ΔADB ~ ΔCDA, their corresponding sides are in proportion:

`(AD)/(CD) = (DB)/(AD)`

Let AD = x and CD = 2 cm. 

Also, we know: 

BC = BD + DC = 10

⇒ BD = 10 − 2 = 8cm

`x/2 = 8/x`

Cross multiplying:

`x^2 = 16`

`x + sqrt16 = 4`cm

Thus, AD = 4 cm.