Question

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE^2 + BD^2 = AB^2 + DE^2.

Answer 1

Solution: In ∆ ACE; AE² = AC² + CE² ……… (1)

In ∆ DCB; BD² = DC² + CB² ……… (2)

In ∆ ACB; AB² = AC² + CB² ……… (3)

In ∆ DCE; DE² = DC² + CE² ……… (4)

Adding equations (1) and (2), we get;

AE² + BD² = AC² + CE² + DC² + CB² …….. (5)

Adding equations (3) and (4), we get;

AB² + DE² = AC² + CB² + DC² + CE² ………. (6)

On comparing the RHS of equations (5) and(6), we get;

AE² + BD² = AB² + DE² proved