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The line p is parallel to the the side AB of triangle ABC and splits the sides AC and BC in points D and E, respectively. If the area of triangle ABD is m and the area of triangle AEC is n, find the area of ABC.

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It is clear that $P_{BDE}=P_{ADE}$, so $P_{BCD}=P_{AEC}=n$. Hence $$P_{ABC}=P_{ABD}+P_{BCD}=m+n.$$

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ABD and ABE have the same base and height, so must have the same area. Thus ABC area equals m+n

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  • $\begingroup$ Excellent! This looks more straightforward than mine. $\endgroup$
    – Solumilkyu
    Mar 27 '16 at 15:24
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I think visually all you have to see is that $P_{ADE}=P_{BDE}$ (triangles with the same base between the same pair of parallel lines must have the same area.)

Let's set $AE\cap DB:=X$ for ease of notation

Since $P_{ADE}=P_{BDE}$, we have $P_{ADX}=P_{BEX}$ by subtracting $P_{DXE}$ from both sides of the original equation.

Notice by disection and associativity on multiplication: $$P_{ABC}=P_{CDE}+P_{DXE}+P_{ADX}+P_{BEX}+P_{ABX}$$ $$=P_{ABX}+P_{ADX}+P_{CDE}+P_{DXE}+P_{BEX}$$

We further know that $P_{ABD}=P_{ABX}+P_{ADX}$ thus the equality becomes: $$=P_{ABD}+P_{CDE}+P_{DXE}+P_{BEX}$$

Since $P_{ADX}=P_{BEX}$, we have $$=P_{ABD}+P_{CDE}+P_{DXE}+P_{ADX}$$

Finally, $P_{AEC}=P_{CDE}+P_{DXE}+P_{ADX}$ due the disection and hence: $$=P_{ABD}+P_{AEC}=m+n$$

This is a more explicit (though confusing due to all the letters) way of seeing it but it's really up to you. The others work great too! Hope that helps

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