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Question Number 169092 by Shrinava last updated on 24/Apr/22

	Answered by mr W last updated on 24/Apr/22

	Commented by mr W last updated on 24/Apr/22

	${B E}^{2} = a^{2} + b^{2} - 2 a b \cos (π - \frac{π}{2} - θ)$ ${B E}^{2} = a^{2} + b^{2} - 2 a b \sin θ = {(\sqrt{2} c)}^{2} . . . (i)$ $r e p l a c e θ w i t h - θ w e g e t$ ${B D}^{2} = a^{2} + b^{2} + 2 a b \sin θ = {(\sqrt{2} d)}^{2} . . . (i i)$ $(i) + (i i) :$ $2 (a^{2} + b^{2}) = 2 c^{2} + 2 d^{2}$ $\Rightarrow a^{2} + b^{2} = c^{2} + d^{2} p r o v e d ✓$ $i t c a n a l s o b e p r o v e d t h a t t h e b l u e$ $s q u a r e a n d t h e g r e e n s q u a r e a l w a y s$ $t o u c h e a c h o t h e r a s s h o w n .$ $\sin (α + \frac{π}{4}) = \frac{a \cos θ}{\sqrt{a^{2} + b^{2} - 2 a b \sin θ}}$ $\sin (β + \frac{π}{4}) = \frac{a \cos θ}{\sqrt{a^{2} + b^{2} + 2 a b \sin θ}}$ ${(\sqrt{2} a)}^{2} = c^{2} + d^{2} + 2 c d \sin (α + β) ?$ $2 a^{2} = c^{2} + d^{2} - 2 c d \cos (α + \frac{π}{4} + β + \frac{π}{4})$ $b^{2} - a^{2} \sin^{2} θ = \sqrt{{(a^{2} \sin^{2} θ + b^{2})}^{2} - 4 a^{2} b^{2} \sin^{2} θ}$ $b^{4} + a^{4} \sin^{4} θ = {(a^{2} \sin^{2} θ + b^{2})}^{2} - 2 a^{2} b^{2} \sin^{2} θ$ $0 = 0 ✓$
	Commented by Tawa11 last updated on 24/Apr/22

	$Great sir .$
	Commented by Shrinava last updated on 24/Apr/22

	$Perfect dear sir thank you$
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