In a quadrilateral, angle A+angle D=90'. …
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Sia ? 6 years, 5 months ago
Given: In quadrilateral ABCD,
{tex}\angle A + \angle D = 90 ^ { \circ }{/tex}
AC and BD are joined
To prove: AC2 + BD2 = AD2 + BC2
Construction: Produce AB and DC to meet at P.
Proof: In {tex}\triangle{/tex}APD,
{tex}\angle A + \angle D = 90 ^ { \circ }{/tex}(given)
{tex}\therefore \angle P = 90 ^ { \circ }{/tex} {tex}\left( \because \angle A + \angle P + \angle D = 180 ^ { \circ } \right){/tex}
Now in right {tex}\triangle A C P , \angle A P D = 90 ^ { \circ }{/tex}
AC2 = PA2 + PC2 ....(i)
(Pythagoras Theorem)
and in {tex}\triangle{/tex}BPD
BD2 = PB2 + PD2
Adding (i) and (ii)
AC2 + BD2 = PA2 + PC2 + PB2 + PD2
= (PA2 + PD2) + (PC2 + PB2)
= AD2 + BD2
({tex}\therefore{/tex} In right {tex}\triangle{/tex}APD, PA2 + PD2 = AD2 and similarly PC2 + PB2 = BD2)
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