Question

In the given figure, XY || AC and XY divides ΔABC into two regions, equal in area. Show that `(AX)/(AB) = ((2 - sqrt(2)))/2`.

Answer 1

Given:

In triangle ABC, XY || AC, with X on AB and Y on BC.

The segment XY divides ΔABC into two regions of equal area.

Show `(AX)/(AB) = ((2 - sqrt(2)))/2`.

Step-wise calculation:

1. Because XY || AC, triangles BXY and BAC are similar (corresponding vertices B→B, X→A, Y→C).

2. Let AB = 1 (work with a unit length along AB); let BX = t, so AX = 1 – t.

3. Similarity gives `(Area(ΔBXY))/(Area(ΔBAC)) = ((BX)/(BA))^2 = t^2`.

4. The condition “XY divides ΔABC into two equal areas”

⇒ Area(ΔBXY) = `1/2` × Area(ΔBAC). 

Therefore `t^2 = 1/2`.

5. So `t = 1/sqrt(2)`.

Hence `(AX)/(AB) = 1 - t`

= `1 - 1/sqrt(2)`

6. Simplify `1 - 1/sqrt(2)`: 

`1 - 1/sqrt(2) = sqrt(2)/sqrt(2) - 1/sqrt(2)` 

= `(sqrt(2) - 1)/sqrt(2)`

= `(2 - sqrt(2))/2`

`(AX)/(AB) = ((2 - sqrt(2)))/2`. This is the required result.