Question

in figure,XY ll AC and
XY divides ABC into
two equal areas.
determine AX:AB​

Answer 1

Answer:

AX/AB = (2-√2)/2

Step-by-step explanation:

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, ΔABC ~Δ 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  = √2/1

XB/AB = 1/√2

1- XB/AB = 1-1/√2

(AB - XB)/AB =(√2 -1)/√2

Therefore, AX/AB = (√2 -1)/√2

Answer 2

AX:AB=CY:CB

(by property of pararel lines and transversal)

hope it helps you