Question

Answer with proper explantion.

Answer 1

Hey, Zaransha.

Here we go sweetie,


So as ar(AXYC)=ar(ΔBYX)

adding ar(ΔBYX) both sides,


ar(ΔABC)= 2 ar(ΔBYX)

Therefore,

$\frac{ ΔBYX }{ ΔABC } = \frac{1}{2}$
-----(i)


From Area theorem we know that,

$\frac{ ΔBYX }{ΔABC } = \frac{ {BX }^{2} }{ {AB }^{2} }$
-----(ii)



Using (i) and (ii)
$\frac{ {BX}^{2} }{ {AB}^{2} } = \frac{1}{2} \\ \frac{ {(AB -AX)}^{2} }{ {AB}^{2} } = \frac{1}{2} \\ \frac{AB -AX}{AB} = \frac{1}{ \sqrt{2} } \\ \\ \frac{AB}{AB} - \frac{AX}{AB} = \frac{1}{ \sqrt{2} } \\ \\ \\ \frac{AX}{AB} = \frac{1}{ \sqrt{2} } - 1 = \frac{1 - \sqrt{2} }{ \sqrt{2} }$


After rationalizing the denominator you'll get,

$\frac{AX}{AB} = \frac{ \sqrt{2} - 2}{ \sqrt{2} }$

Hence proved.