# Diagonals of convex $\square ABCD$ meet at $O$. Show that $|\triangle AOB| = |\triangle COD|$ if and only if $BC \parallel AD$

Let $$O$$ be the intersection of the diagonals in a convex quadrilateral $$\square ABCD$$, Show that $$|\triangle AOB| = |\triangle COD|$$ (that is, the areas are equal) if and only if $$BC$$ is parallel to $$AD$$.

I have tried to move the corner $$C$$ on $$AC$$. When it is far away, $$\triangle OCD$$ has a big area, and when it is at $$O$$ the area have to be $$0$$. So, somewhere in between those spots the area of $$\triangle AOB$$ must be equal to the area of $$\triangle COD$$. Also, there should be two similar triangles then, but how do I show that the areas are the same only if they are parallel?

• @user1551 It can, how? Because the question I'm working on just states |$\triangle$$AOB| = |\triangle$$COD$| if and only if $BC$ and $AC$ are paralell, if I could show it's not true it would be great! – Alli Henne Nov 1 '18 at 9:27
• Hint: Consider $\triangle ABC$ and $\triangle BCD$. – Blue Nov 1 '18 at 9:43

Suppose that areas of $$\triangle COD$$ and $$AOB$$ are equal. It means areas of triangles $$\triangle BCD$$ and $$\triangle BCA$$ are also equal. If you pick $$BC$$ as the base, it means that heights of triangles $$\triangle BCD$$ and $$\triangle BCA$$ with respect to the base $$BC$$ must be equal. These two heights actually represent distances of points $$A$$ and $$D$$ from line BC. Because these distances are the same, it means that points $$A$$ and $$D$$ must be on a line parallel with BC: $$AD\parallel BC$$.