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Given:triangle ABC. In the side AB, we choose point D. From this point D, we draw a line DF such that intersect side AC and line DE such that intersect side BC. If DF//BC, DE//AC, and the area of BDE = p times area of ADF, what is the ratio of area CEF and ABC?

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We give a scaling argument. You can change it to a more conventional argument, using expressions for the area of a triangle.

Triangles $ADF$ and $DBE$ are similar, and each is similar to $\triangle ABC$. The areas of $\triangle ADF$ and $DBE$ are om the ratio $1$ to $p$. With the right measure of area, their areas can be taken to be $1$ and $p$.

So sides $AD$ and $DB$ are in the ratio $1$ to $\sqrt{p}$. (For recall that scaling linear dimensions by scaling factor $\lambda$ scales areas by $\lambda^2$.) This means that $AD$ is to $AB$ as $1$ to $1+\sqrt{p}$.

Hence $\triangle ABC$ has area $(1+\sqrt{p})^2$, that is, $1+2\sqrt{p}+p$.

It follows that the parallelogram $FDEC$ has area $2\sqrt{p}$. So triangle $CEF$ has area $\sqrt{p}$. The ratio of its area to the area of the whole triangle $ABC$ is $\sqrt{p}$ to $(1+\sqrt{p})^2$.

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thanks for the explanation, Andre. But I still don't understand why the ratios of AD and DB is 1 to sqrt(p) ? where sqrt(p) come from? –  akusaja Aug 14 '13 at 14:27
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Let $AD=x$ and $DB=\lambda x$. Let the height of $\triangle ADF$ be $h$. Then by similarity the height of $\triangle DBE$ is $\lambda h$. So area of $\triangle ADF$ is $(1/2)xh$ and the area of $\triangle ADF$ is $(1/2)(\lambda x)(\lambda h)=\lambda^2(1/2)(xh)$. That means the ratio of $(1/2)xh$ to $\lambda^2(1/2)xh$ is $1$ to $p$. So $1$ to $\lambda^2$ is same as $1$ to $p$, giving $\lambda=\sqrt{p}$. (I have done it the ugly way!) –  André Nicolas Aug 14 '13 at 14:54
    
My last question is 'How can you know that the area of triangle ABC is (1+sqrt p)^2? Because we didn't know the measure of its altitude. Thanks –  akusaja Aug 14 '13 at 17:10
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This triangle is similar to the two little ones at the bottom. The base of $\triangle ABC$ is $1+\sqrt{p}$ times the base of $\triangle ADF$. So by similarity the height of $\triangle ABC$ is $1+\sqrt{p}$ times the height of $\triangle ADF$. Now I expect you can see the result. It is basically the same argument as the one one comment back. –  André Nicolas Aug 14 '13 at 17:16

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