Question

please send answer for this question quickly​

Answer 1

in triangle BXY and triangle BAC

$< b = < b (commom)$

$< x = < y (corrsponding angles)$

therefore BAC similar to BXY

( AA) CONDITION

$bxy \div bac = bx ^{2} \div ba {}^{2}$

$1 \div \sqrt{2 } = bx \div ab$

subtracting 1 on both sides

we get,

$1 - 1 \div \sqrt{2} = 1 - bx \div ba$

$\sqrt{2} - 1 \div \sqrt{2} = ba - bx \div ba$

$\sqrt{2} ( \sqrt{2} - 1) \div 2 = ax \div ab$

$2 - \sqrt{2} \div 2 = ax \div ab$

hence proved
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