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

$\huge\ { \fbox{ \tt{ Answer}}}$

REF.Image

2 Area (xy) = Area (ABC)

In ΔABC & ΔXBY

∠ABC=∠XBY(common)

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

ΔABC∼ΔXBY (AA similarity)

(XBAB)2=Area(ΔBXY)Area(ΔABC)=2

(AB)2=2(XB)2    AB=2×B

AX=AB−XB=AB−2AB=(22−1)AB

ABAX=22−1

Answer 2

$\mathtt \green {Given : \:}$

$in \: ∆ABC \\ •\:XY \: ll \: to \: AC \\ •\: ar(∆BXY) = ar (∆AXYC)$

$\mathtt \green {To fine : \:}$

$\frac{AX}{AB}$

$\mathtt \green {Proof ;\:}$

$In \:∆ABC\: and \:∆XBY$

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