Question

the line segment XY is parallel to side AC of ∆ABC and it divides the traingle into two parts of equal areas find the ratio AX/AB​

Answer 1

Answer:

It is given that :-

ar(∆XBY) = ar(trap. XACY)

ar(∆XBY)+ar(∆XBY)= ar(trap.XACY)+ar(∆ABC)

2ar(∆XBY) = ar(∆ABC) ...(1)

Now in ∆ ABC and ∆XBY we have,

angle ABC= angle XBY

angle BAC= angle BXY

SO ∆ ABC ~ ∆XBY

So

ar(∆ABC)/ar(XBY)= (AB/BX)^2

(AB/BX)^2= 2/1 [using 1]

AB/BX= √2/1

BX/AB= 1/√2

1-BX/AB= 1-1/√2

(AB-BX)/AB= (√2-1)/√2

AX/AB=(√2-1)/√2

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