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The diagram can be viewed from this picture.

In the diagram the quadrilateral ABCD has point M on AB and point N on DC. We are given that $\frac{AM}{AB}=\frac{NC}{DC}$. Also we are given that the line segments AN and DM intersect at P, and MC and NB intersect at Q. How do we prove that the area of quadrilateral MNPQ equals the sum of the areas of triangle APD and triangle BQC.

My progress: The only thing that stands out is that the whole entire problem is most likely based on taking ratio of areas. I've played around with this for like 30 minutes to an hour and got nothing. No lines are parallel so I dont see what else can be done.


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This is equivalent to showing that the areas of $\Delta ABN$ and $\Delta CDM$ precisely add up to the area of $ABCD$. Now let $t = AM/AB = CN/CD$. Then the area of $ABN$ is linear in this parameter $t$ and goes from $\textrm{area}(ABC)$ to $\textrm{area}(ABD)$ for $t \in [0,1]$. (The base $AB$ is constant while the height changes linearly in $t$.) Likewise the area $CDM$ is linear in $t$ and goes from $\textrm{area}(ACD)$ to $\textrm{area}(BCD)$. Now add these linear functions and see what you get.

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I dont exactly understand what you mean by express as a function. Can you please tell? thanks very much! – ahuang Apr 15 '12 at 12:17
@AlanHuang The area of $ABN$ depends on the ratio $AM/AD$ or in other words, it is a function of that ratio. – WimC Apr 15 '12 at 12:32
Can you please go into more detail. I do not see how you can set a function for height in terms of $t$. Any help is greatly appreciated i have been stuck on this problem for a while! Thanks! – ahuang Apr 15 '12 at 17:10
For $ABN$ the area goes linearly from area$(ABC)$ at $t=0$ to area$(ABD)$ at $t=1$ and is therefore given by $(1-t)\cdot\textrm{area}(ABC) + t \cdot\text{area}(ABD)$. Similarly an explicit expression can be found for $CDM$. – WimC Apr 15 '12 at 20:09

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