Question

In quadrilateral ABCD, angle B=90°,AD square=AB square+BC square+CD square,then prove that angle ACD=90°.

Answer 1

As ABC is a right angled triangle, with AC the hypotenuse, we have that:

$AC^{2}  = AB^{2} + BC^{2}$

Therefore, we have:

$AD^{2} = AB^{2} + BC^{2} + CD^{2}$

⇒ $AD^{2} = AC^{2} + CD^{2}.$

Hence, as Δ ACD satisfies Pythagoras' theorem, with AD the hypotenuse, ∠ACD = 90°, as required.