In Fig. 3, ACB = 90⁰ and CD AB, prove that CD² = BD × AD.

Or

If P and Q are the points on side CA and CB respectively of ∆ ABC, right angled at C, prove that (AQ² + BP²) = (AB² + PQ²)

Question

In Fig. 3, ACB = 90⁰ and CD AB, prove that CD 2 = BD × AD. Or If P and Q are the points on side CA and CB respectively of ∆ ABC, right angled at C, prove that (AQ 2 + BP 2 ) = (AB 2 + PQ 2 )

Philoid · Accepted Answer

By Pythagoras theorem

In ∆ ABC

AB² = AC² + BC² …(1)

In ∆ ACD

AC² = AD² + DC² …(2)

In ∆ BCD

BC² = BD² + CD² …(3)

Add eq (2) and eq (3)

AC² + BC² = AD² + DC² + BD² + CD²

AB² = AD² + 2CD² + BD²

AB² - AD² - BD² = 2CD²

(AB – AD)(AB + AD) – BD² = 2CD²

BD(AB + AD – BD) = 2CD²

BD(2AD) = 2CD²

CD² = BD × AD

Or

LHS = AQ² + BP²

= (CQ² + CA²) + (CP² + CB²)

= (CQ² + CP²) + (CA² + CB²)

= PQ² + AB²

= RHS

Hence proved.

In Fig. 3, ACB = 90⁰ and CD AB, prove that CD2 = BD × AD.OrIf P and Q are the points on side CA and CB respectively of ∆ ABC, right angled at C, prove that (AQ2 + BP2) = (AB2 + PQ2)

In Fig. 3, ACB = 90⁰ and CD AB, prove that CD² = BD × AD.

Or

If P and Q are the points on side CA and CB respectively of ∆ ABC, right angled at C, prove that (AQ² + BP²) = (AB² + PQ²)