Question

In triangle ABC XY is parallel to BC and XY divides the triangle inti two parts of equal area find BX each to AB

Answer 1

pls mark it as brainliest answer.

Answer 2

AX/XB = 1 /√2

Step-by-step explanation:

Given:

In ΔABC, XY || BC and area of ΔAXY = area of quadrilateral XYCB

⇒ ar (ΔABC) = 2ar (ΔAXY)  

⇒ $\frac{ar (\triangle AXY)}{ar (\triangle ABC)} =\frac{1}{2}$                   (1)

XY || BC and AB is a transversal.

⇒ ∠AXY = ∠ABC                                         (2)

So, In ΔABC and ΔAXY,

∠A = ∠A            [common angle]

∠AXY = ∠ABC            [from equation (2)]

⇒ ΔABC ~ ΔAXY

⇒ $\frac{ar(\triangle AXY)}{ar(\triangle ABC)} =\frac{(AX)^2}{(AB)^2}$

⇒ $\frac{1}{2} =\frac{(AX)^2}{(AB)^2}$

⇒$\frac{AX}{AB}= \sqrt{\frac{1}{2} }$

⇒$\frac{AX}{AB}=\frac{1}{\sqrt{2} }$