Question

The diagonals of a convex ؈BCD intersect at right angles. Prove that AB²+CD²=AD²+BC².

Answer 1

Using Pythagoras theorem :
AO² + OB² = AB² ...............(i) [from ∆AOB right angled triangle]
DO² + OC² = CD² …..........(ii) [from ∆COD right angled triangle]
AO² + OD² = AD² …..........(iii) [from ∆AOD right angled triangle]
BO² + OC² = BC² ….........(iv) [from ∆BOC right angled triangle]
Adding Equation(i) and Equation(ii)
AB² + CD² = AO² + OB² + DO² + OC² …..(v)

Adding Equation(iii) and Equation(iv)
AD² + BC² = AO² + OB² + DO² + OC² …...(vi)

we can see that the RHS of equation (v) and (vi) are similar
So we can say that
AB² + CD² = AD² + BC²