Use the fact that similar triangles have area scaled by the square of the factor of similarity. The sum of the areas of the shaded regions is $S = (1/4-1/16+1-9/16)S_{ABD}+(1/16+9/16-1/4)S_{ACD}=\frac{10}{16}S_{ABD}+\frac{6}{16}S_{ACD}$. The sum of the unshaded areas is $\bar S = \frac{6}{16}S_{ABD}+\frac{10}{16}S_{ACD}$. If you equal them you get $S_{ABD}=S_{ACD}$ which gives $BD=DC$.

benimath wrote: Use the fact that similar triangles have area scaled by the square of the factor of similarity. The sum of the areas of the shaded regions is $S = (1/4-1/16+1-9/16)S_{ABD}+(1/16+9/16-1/4)S_{ACD}=\frac{10}{16}S_{ABD}+\frac{6}{16}S_{ACD}$. The sum of the unshaded areas is $\bar S = \frac{6}{16}S_{ABD}+\frac{10}{16}S_{ACD}$. If you equal them you get $S_{ABD}=S_{ACD}$ which gives $BD=DC$. @Bunny da : Are we given that those small triangles are similar? Or any such extra information about those small triangles?

Umm, yeah, you're given that those lines are parallel (though that wasn't explicitly stated in the paper).

easiest rmo problem ever .

yea it was just very easy just had to do apply thales theorem and area of trapezium and triangle. which easily gives $ BD = DC $

What is thales theorem ??

basic proportionality theorem.

Okay yeah. Didn't know that name for it.

It is not given that the lines are parallel. -_-

I think that you can call them parallel WLOG.

how can you do that they are certainly not given parallel

If you don't assume that they are parallel, the problem is impossible.

U r sure?

Prove that they have to be parallel to have alt areas equal.

Just use Area of triangle = 1/2 bh and area of trapezium = (a+b)h/2