Question

In the given figure, the line segment XY is
parallel to side AC of AABC and it divides
the triangle into two equal parts of equal
area. Find the ratio AXIAB.​

Answer 1

Answer:

Given: ∆ABC in which XY || AC and ar(BXY)=ar(AXYC)

To Prove : AX/AB

Proof : In ∆ ABC and ∆BXY

* ang BYX = ang C

* ang BXY = ang A

* ang B = ang B

.•.∆BXY ~ ∆ABC. { AAA}

.•.ar∆ABC/ ar∆BXY. = AB²/BX²

=>. 2ar∆ BXY/ ar∆BXY. = AB²/BX²

2 = AB²/BX²

2. = { AB/BX }²

√2 = AB/BX

√2 = AB/AB-AX

AB/AB - AX/AB. = 1/√2

1 - 1/√2 = AX/AB

AX/AB = √2 - 1/√2

AX/AB = 2 - √2/2. .....

Answer 2

Answer:

$\huge\mathfrak\red{in attachment}$

Angle A = Angle C

by [ corresponding angle]

hope it's helpful for you .