Question

(b) in a triangle abc XY is parallel to BC and it divides triangle ABC into two parts of equal area. Prove that BX/AB = √2-1/√2​

Answer 1

Answer:

Hope you get it...

Area in the answer could be directly equated with square of sides as a theorem states......

Step-by-step explanation:

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Answer 2

Given :  XY is parallel to BC'

To prove : $\frac{BX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}$

Proof :

it divides triangle ABC into two parts of equal area.

i.e

In traingle ABC and AXY

$ar(\bigtriangleup ABC)=2 ar(\bigtriangleup AXY)\\\\\frac{ar(\bigtriangleup AXY)}{ar(\bigtriangleup ABC)}=\frac{1}{2}\\\\\angle A = \angle A \\\\\angle AXY = \angle ABC$

therefore by AA similarity

$\bigtriangleup AXY\sim \bigtriangleup ABC\\\\\frac{ar(\bigtriangleup AXY)}{ar(\bigtriangleup ABC)}=\frac{AX^2}{AB^2}$

ratios of the corresponding sides of the similar triangle

now comparing

$\frac{ar(\bigtriangleup AXY)}{ar(\bigtriangleup ABC)}=\frac{AX^2}{AB^2} and \frac{ar(\bigtriangleup AXY)}{ar(\bigtriangleup ABC)}=\frac{1}{2}\\\\\frac{AX^2}{AB^2}=\frac{1}{2}\\\\\frac{AX}{AB}=\frac{1}{\sqrt{2}}\\\\\frac{AB-AX}{AB}=\frac{1}{\sqrt{2}}\\\\\frac{AB}{AB}-\frac{BX}{AB}=\frac{1}{\sqrt{2}}\\\\1-\frac{BX}{AB}=\frac{1}{\sqrt{2}}\\\\\frac{BX}{AB}=1-\frac{1}{\sqrt{2}}\\\\\frac{BX}{AB}=\frac{\sqrt{2}-1}{\sqrt{2}}$

hence proved

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