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: BC = 10 cm, CD = 2 cm

To prove: ΔADB ~ CDA

To find: AD

In ΔCAB and ΔADB

∠CAB = ∠ADB = 90°

∠CBA = ∠DBA   ....(Common)

∴ ΔCAB ~ ΔADB ....(AA similarity) (i)

In ΔCAB ~ ΔCDA

∠CAB = ∠CDA = 90°

∠ACB = ∠DCA   ...(Common)

∴ ΔCAB ~ ΔCDA   ...(AA similarity) (ii)

By equations (i) and (ii),

ΔADB ~ ΔCDA,

If two triangles are similar to the same triangle, then they are similar to each other.

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

AD² = CD × BD

= 2 × 8

= 16

AD = 4 cm