Question

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In a quadrilateral ABCD , angleB - 90 degree - angleD. prove that
2AC^2-BC^2=AB^2+AD^2+DC^2


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Answer 1

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Answer 2

$\underline\bold{\huge{SOLUTION \:\: :}}$

$\underline\bold{GIVEN\:\: :}$

ABCD is a quadrilateral.

<B = <D = 90°

$\underline\bold{TO\:\:PROVE\:\: :}$

2 AC² - BC² = AB² + AD² + DC²

$\underline\bold{CONSTRUCTION\:\: :}$

Join diagonal AC.

$\underline\bold{PROOF\:\: :}$

In ∆ ABC, <B = 90°.

By Pythagoras Theorem,

AC² = AB² + BC² ............. (i)

Again, in ∆ ADC, <D = 90°.

AC² = AD² + DC² ............ (ii)

Adding (i) and (ii), we get :

AC² + AC² = AB² + BC² + AD² + DC²

=> 2 AC² - BC² = AB² + AD² + DC²