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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?

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  • $\begingroup$ @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! $\endgroup$ – Alli Henne Nov 1 '18 at 9:27
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    $\begingroup$ Hint: Consider $\triangle ABC$ and $\triangle BCD$. $\endgroup$ – Blue Nov 1 '18 at 9:43
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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: $AD\parallel BC$.

Reverse the order of sentences and you will prove the opposite statement.

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