Well done Arun from National Public School, Bangalore, India, some quality thinking in devising this solution.

We are given a triangle ABC , and are required to draw a line DE parallel to CB such that it divides the triangle into 2 portions of equal areas.

The area of triangle ABC is double the area of AED.

But, ADE and ACB are similar triangles because DE is parallel to CB

We also know that the ratio of the areas of the two similar triangles is equal to the ratio of the squares of corresponding sides.

Which means that the line ratio AD:AC must be

1 : \sqrt{2}

The problem becomes : how to locate D to achieve this ratio.

A square of side length 1 has a diagonal length of

\sqrt{2}

or, put another way, an isosceles right-angled triangle has a hypotenuse $\sqrt{2}$ times bigger than the other sides.

Here is a construction to achieve this required ratio.

X is any suitable point on AD

ZX is perpendicular to AC, and ZX is equal in length to AX.

So AXZ is an isosceles right-angled triangle.

By sweeping an arc centre A from X to AZ at N, AN is made equal to AX

AN to AZ is now in the required ratio.

Drawing from N parallel to ZC the point D is reached.

Because AND and AZC are similar triangles, AD and AC are in the required ratio.

Excellent and simple!