Question

In the following Figure ∠ACB= 90° and CD ⊥ AB, prove that  CD²  = BD × AD

Answer 1

Given that: CD ⊥ AB 

∠ACB = 90° 

To prove : CD² = BD × AD 

Using pythagoras Theorem in Δ ACD 

AC² = AD² +CD² ..........(1)

Using pythagoras Theorem in ΔCDB

CB² = BD²+CD² .....(2) 

Similarly in ΔABC, 

As AB = AD + DB 

Since ,AB = AD +BD ......(4)

Squaring both sides of equation (4), we get

(AB)² = (AD+BD)²

Since, AB² = AD² +BD² +2×BD×AD

From equation (3) we get 

AC² +BC²=AD²+BD²+2 × BD × AD 

Substituting the value of AC² from equation (1) and the value  of BC² from equation  (2), we get

AD² +CD² +BD² + CD² +AD² +BD² +2×BD×AD

Since ,2 CD² = 2 × BD × AD

Hence , CD² = BD × AD