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 $\triangle ABC$ ?

  • $\begingroup$ @HarishChandraRajpoot: For all your trivial edits going forward, please remember: It's $\parallel$. $\endgroup$
    – Blue
    Jul 9, 2015 at 21:04

2 Answers 2


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 $(1+\sqrt{p})^2$.

  • $\begingroup$ 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? $\endgroup$
    – akusaja
    Aug 14, 2013 at 14:27
  • 1
    $\begingroup$ 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!) $\endgroup$ Aug 14, 2013 at 14:54
  • $\begingroup$ 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 $\endgroup$
    – akusaja
    Aug 14, 2013 at 17:10
  • 1
    $\begingroup$ 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. $\endgroup$ Aug 14, 2013 at 17:16

The quadrilateral $CEDF$ is a parallelogram, and therefore $CE=DF=d$, $CF=DE=e$, and $\triangle CEF \equiv \triangle DEF$.

Triangle 2

Suppose $h$ is the height of parallelogram $CEDF$ and $\ell$ is the height of $\triangle ADF$. Then, $h$ will also be the height of $\triangle BDE$ and $\triangle CEF$.

$$\therefore \ \text{Area of }\triangle CEF = \text{Area of }\triangle DEF = \frac12 \text{ Area of $CEDF$ parallelogram} = \frac12 dh$$

As shown in the diagram (use your knowledge on parallelograms), you may find $\triangle BDE$ and $\triangle ADF$ are similar. Thus, their corresponding side lengths have the same ratio:

$$ \frac{BE}{DF} = \frac{BD}{DA} = \frac{DE}{AF} = \frac{c}{d} = \frac{b}{a} = \frac{e}{f} \tag1$$

Suppose areas of $\triangle BDE$ and $\triangle ADF$ are $A_{\triangle BDE}$ and $A_{\triangle ADF}$, respectively. Given that:

$$\frac{A_{\triangle BDE}}{A_{\triangle ADF}} = p \ \Rightarrow \ A_{\triangle ADF} = \frac{1}{p} A_{\triangle BDE} \tag2$$

$$\therefore \ \frac{\frac12ch}{\frac12d\ell} = \frac{ch}{d\ell} = p \ \Rightarrow \ \frac{c}{d} = \frac{\ell}{h}p \tag 3$$

From the equations $(1)$ and $(2)$, $\frac{b}{a} = \frac{\ell}{h}p$ and $\frac{e}{f} = \frac{\ell}{h}p$, as well.

You may also find $\triangle CGF$ and $\triangle AFH$ are similar as well. Thus, their corresponding side lengths have the same ratio: $$ \frac{AF}{CF} = \frac{AH}{FG} = \frac{f}{e} = \frac{\ell}{h} \ \Rightarrow \ \frac{e}{f} = \frac{h}{\ell} $$ However, we previously found $\frac{e}{f} = \frac{\ell}{h}p$ $$\therefore \frac{\ell}{h}p = \frac{h}{\ell} \ \Rightarrow \ \frac{h^2}{\ell^2} = p \ \Rightarrow \ \frac{h}{\ell} = \sqrt{p} \tag4$$

$$\frac{A_{\triangle BDE}}{A_{\triangle CEF}} = \frac{\frac12ch}{\frac12dh} = \frac{ch}{dh} = \frac{c}{d} = \frac{\ell}{h}p = \frac{p}{\sqrt{p}} \ \Rightarrow \ \therefore \ A_{\triangle BDE} = \frac{p}{\sqrt{p}} A_{\triangle CEF} \tag5$$

From the equation $(2)$: $$A_{\triangle ADF} = \frac{1}{p} A_{\triangle BDE} = \frac{1}{p} \frac{p}{\sqrt{p}} A_{\triangle CEF} = \frac{1}{\sqrt{p}} A_{\triangle CEF} \tag6$$

If the area of $\triangle ABC$ is $A_{\triangle ABC}$, then:

$$A_{\triangle ABC} = A_{\triangle ADF} + A_{\triangle BDE} + 2 \times A_{\triangle CEF} = \frac{1}{\sqrt{p}} A_{\triangle CEF} + \frac{p}{\sqrt{p}} A_{\triangle CEF} + 2 \times A_{\triangle CEF} \\= \left(\frac{1}{\sqrt{p}} + \frac{p}{\sqrt{p}} + 2 \right)A_{\triangle CEF} $$

$$\therefore \ \frac{ A_{\triangle ABC}}{A_{\triangle CEF} } = \frac{\sqrt{p}}{p}(1+p) + 2$$


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