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**In a quadrilateral ABCD, ∠B = 90° and ∠D = 90°.**

Prove that: 2AC^{2} - AB^{2} = BC^{2} + CD^{2} + DA^{2}

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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,

AC^{2} = AB^{2} + BC^{2}

⇒ AB^{2} = AC^{2} - BC^{2 } ....(i)

In ΔADC, using Pythagoras theorem,

AC^{2} = AD^{2} + DC^{2} ....(ii)

LHS = 2AC^{2} - AB^{2}= 2AC^{2} - ( AC^{2} - BC^{2} ) .....[ From(i) ]

= 2AC^{2} - AC^{2} + BC^{2}= AC^{2} + BC^{2}

= AD^{2} + DC^{2} + BC^{2 } ....[ From(ii) ]

= RHS

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