(Q11) ABC is a right triangle right angled at B. Let D and E be any points on AB and BC respectively. Prove that AE2 + CD2 = AC2 + DE2.

Solution :

ABCDE

Given : In ∆ABC; ∠B = 90°

D and E are points on AB and BC.

R.T.P :AE2 + CD2 = AC2 + DE2

Proof : In ∆BCD, ∆BCD is a right triangle right angled at B.

∴ ( )2 + BC2 = ( )2.....(1)

[∵ Pythagoras theorem states that hypotenuse2 = side2 + side2]

In ∆ABE; ∠B = 90°

( )2 = AB2 + ( )2 .....(2)

Adding (1) and (2), we get

( )2 + BC2 + AB2 + ( )2 = ( )2 + ( )2

( 2 + 2) + (AB2 + BC2) = ( )2 + ( )2

DE2 + AC2 = CD2 + AE2

[∵ (i) In ∆DBE, ∠B = 90° and DE2 = BD2 + BE2

(ii) In ∆ABC, ∠B = 90° and AB2 + BC2]