Question

In the given figure, line segment XY is parallel to side AC of triangle ABC and it divides the triangle into two parts of equal area. Prove that AX:AB =(2-_/2):2.

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Answer 1

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Answer 2

Answer:

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In the given figure, line segment XY is parallel to side AC of triangle ABC and it divides the triangle into two parts of equal area. Prove that AX : AB = √2 -1 / √2

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2 Area (xy) = Area (ABC)

In ΔABC & ΔXBY

∠ABC=∠XBY(common)

∠ACB=∠XYB (since XY||AC, angles are equal)

ΔABC∼ΔXBY (AA similarity)

$( \frac{AB}{XB})^{2} = \frac{Area(ΔABC)}{Area(ΔBXY)} = 2 \\ (AB)^2=2(XB)^2    AB \: = \sqrt{2}×B \\ AX=AB−XB=AB− \frac{AB}{\sqrt{2} } = \: ( \frac{ \sqrt{2} - 1}{ \sqrt{2} } )AB \\ \frac{AX}{AB} = \frac{ \sqrt{2} - 1}{ \sqrt{2} }$