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I got a question, and have been unable to solve it. I asked a lot of people, but no one can answer.

Let there be a Quadrilateral $ABCD$. $M$ and $N$ are the midpoints of $AD$ and $BC$ respectively. Prove that $\text{Area}(AND)+\text{Area}(BMC)=\text{Area}(ABCD)$.


My attempt:

Let $AN$ and $BM$ intersect at $E$, and let $DN$ and $CM$ intersect at $F$.

This is possible only if $\text{Area}(AEB)\text{Area}(CFD)=\text{Area}(MENF)$.

I am unable to go farther. I tried to construct diagonals and use the Midpoint Theorem, but it didn't work, and I can't find anywhere to use congruency or similarity.

Please help.

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$$ \begin{align} \text{Area}(ABN)+\text{Area}(CDN) &={1\over2}BN\cdot AK+{1\over2}CN\cdot DL \\&={1\over2}BN(AK+DL) \\&=BN\cdot MH \\&={1\over2}BC\cdot MH \\&=\text{Area}(BCM). \end{align} $$

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  • $\begingroup$ How do you $\Delta ABN$ to be $\frac12BN{}\cdot{}AH$? Shouldn't it be $AK$? $\endgroup$ – MalayTheDynamo Apr 25 '18 at 13:32
  • $\begingroup$ Yes, it was a typo, thank you. I'll correct in a minute. $\endgroup$ – Aretino Apr 25 '18 at 13:35
  • $\begingroup$ Okay, now this answer is officially awesome. +1, and accepting soon if I don't get any more awesome answers. $\endgroup$ – MalayTheDynamo Apr 25 '18 at 13:36

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