Question

The diagonals of quadrilateral ABCD intersect at O. Prove that $\frac{\mathrm{ar}\left(∆\mathrm{ACB}\right)}{\mathrm{ar}\left(∆\mathrm{ACD}\right)}=\frac{\mathrm{BO}}{\mathrm{DO}}$.

Answer 1

We are given the following quadrilateral with O as the intersection point of diagonals

To Prove:

Given ACB and ACD are two triangles on the same base AC

Consider h as the distance between two parallel sides

Now we see that the height of these two triangles ACB and ACD are same and are equal to h

So

Now consider the triangles AOB and COD in which

Therefore,

From equation (1) and (2) we get

Hence prove that