Question

PLZ GIVE ME ANSWER FAST​

Answer 1

Theorem : The ratio of the areas of

two similar triangles is equal to the

square of the ratio of their

corresponding sides.

It is given that,

In Triangle ABC, XY parallel to AC and

XY divides the triangle into two parts of

equal area.

The triangle ABC is attached with this

answer

ar (ABC) = 2 ar (XBY)

We have XY || AC, therefore <BXY =

<A and<BYX = <C (Corresponding

angles)

So, triangle ABC ~ triangle XBY

ar (ABC)/ar (XBY) =(AB/XB)^2

ar (ABC)/ar (XBY) = 2/1 [ since ar

(ABC) = 2 ar (XBY)]

Therefore, (AB/XB )^2= 2/1

AB/XB = v2/1

XB/AB = 1/√2

XB/AB = 1-1/√2

(AB - XB)/AB =(√2 -1)/√2

Therefore, AX/AB = (√2 -1)/√2 = (2-√2)/2