Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $DF\parallel BC$, $DE\parallel AC$, and the area of $\triangle BDE = p$ times area of $\triangle$ADF, what is the ratio of area $\triangle CEF$ and $\triangle ABC$ ?

share|cite|improve this question
@HarishChandraRajpoot: For all your trivial edits going forward, please remember: It's $\parallel$. – Blue Jul 9 at 21:04

1 Answer 1

up vote 2 down vote accepted

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

$\triangle ADF$ and $\triangle DBE$ are similar, and each is similar to $\triangle ABC$. The areas of $\triangle ADF$ and $\triangle DBE$ are of 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 $(1+\sqrt{p})^2$.

share|cite|improve this answer
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
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
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.