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Question
In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
Prove that: 2AC2 - AB2 = BC2 + CD2 + DA2
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Solution
In quadrilateral ABCD, ∠B = 90° and ∠D = 90°.
So, ΔABC and ΔADC are right-angled triangles.
In ΔABC, using Pythagoras theorem,
AC2 = AB2 + BC2
⇒ AB2 = AC2 - BC2 ....(i)
In ΔADC, using Pythagoras theorem,
AC2 = AD2 + DC2 ....(ii)
LHS = 2AC2 - AB2
= 2AC2 - ( AC2 - BC2 ) .....[ From(i) ]
= 2AC2 - AC2 + BC2
= AC2 + BC2
= AD2 + DC2 + BC2 ....[ From(ii) ]
= RHS
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