Question

in the given figure the line segment XY is parallel to the side AC of triangle ABC and it divides triangle into two parts of equal area find the ratio AX BY AB​

Answer 1

Answer:

Step-by-step explanation:

SOLUTION:

GIVEN:

XY || AC

ar(∆BXY) = ar(quad.XYCA)

ar(∆ABC) = 2ar(∆BXY).................(1)

Now, XY || AC & BA is a transversal .

∠BXY = ∠BAC…………………(2)

[corresponding angles]

In ∆ BAC & ∆BXY,

∠BXY = ∠BAC     (Proved above)

∠XBY = ∠ABC      (Common)

∆ BAC ~ ∆BXY    

[By AA similarity criterion]

ar(∆BAC) / ar(∆BXY) = BA² /BX²

[The ratio of the area of two similar triangles is equal to the ratio of the square of any two corresponding sides]

2ar(∆BXY) / ar(∆BXY) = BA² /BX²

2 = BA² /BX²

BA² = 2 × BX²

BA =√2 × BX²

BA = √2 BX

BA = √2 (BA - AX)      [ BX = BA - AX ]

BA = √2BA - √2AX

√2AX = √2BA - BA

√2AX = BA (√2- 1)

AX / AB = (√2- 1) / √2

Hence, proved.

HOPE THIS WILL HELP YOU...