Question

IN triangle ABC XY parallel to AC. XY divides
triangle
ABC into two equal areas
find AX:XB​

Answer 1

Answer:

sorry I don't know the answer of this question I will tell the answer later after checking from net

Answer 2

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 = (2-√2)/2

i hope it will help u...