# 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

ACD = 90° [converse of Pythagoras theorem]

Hence Proved

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