Question

In the given figure, the line segment XY is parallel to AC of triangle ABC and it divides the triangle into two parts of equal area. Prove that AX:AB =
$2 - \sqrt{2} \ratio2$
Solve the problem!! ​

Answer 1

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Since XY ||AC, we have

$\angle \: A = \angle \: BXY \: and \angle \: C = \angle \: BYX \: \: \: \: \: \: \: (corresponding \: angles)$

$\therefore \: \triangle{ABC } \sim \triangle{XBY } \\ \implies \: \frac{ar( \triangle \: ABC )}{ ar( \triangle \: XBY )} = \frac{ {AB }^{2} }{ {XB }^{2} } ............(1)$

But $ar( \triangle \: ABC ) = 2 \times ar( \triangle \: XBY ) \: \: \: \: \: (given) \\ \implies \frac{ar( \triangle \: ABC )}{ ar( \triangle \: XBY )} = 2 \: ...........(2)$

From (1)&(2), we get

$\frac{ {AB }^{2} }{ {XB }^{2} } = 2 \implies \: ( { \frac{AB }{XB } )}^{2} = 2 \\ \implies \: \frac{AB }{XB } = \sqrt{2} \implies \: AB = \sqrt{2} (XB ) \\ \implies \: AB = \sqrt{2}( AB - AX ) \\ \implies \: \sqrt{2}AX = ( \sqrt{2} - 1)AB \\ \implies \frac{AX }{AB } = \frac{( \sqrt{2} - 1) }{ \sqrt{2} } \times \frac{ \sqrt{2} }{ \sqrt{2} } = \frac{(2 - \sqrt{2}) }{2}$

Hence, AX:AB =$(2 - \sqrt{2} ) \ratio2$