Q23 of 95 Page 5

In a quadrilateral, ΔBCD, ∠B = 90°. If AD2= AB2 + BC2 + CD2, prove that ∠ACD = 90°.


Given: ABCD is a quadrilateral and B = 90°


and AD2= AB2 + BC2 + CD2


To Prove: ∠ACD = 90°


In right triangle ∆ABC, using Pythagoras theorem, we have


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (AB)2 + (BC)2 = (AC)2 …(i)


Given: AD2= AB2 + BC2 + CD2


⇒ AD2= AC2 + CD2 [from (i)]


In ∆ACD


AD2= AC2 + CD2


∴ ∠ACD = 90° [converse of Pythagoras theorem]


Hence Proved


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