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I've tried all the possible side splitter and angle bisector theorem stuff and I still can't come up with the correct answer. I even tried some law of cosine and sine stuff, but nothing. Any help would be gladly appreciated. Thanks.

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    Correct answer is 24, I also came to 40 – 2017-02-24
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    I removed my comment because I realised how stupid it was to suggest that DE = BC lol - but I see the mistake I made – 2017-02-24

4 Answers 4

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Observe triangles $ADE$ and $ABC$ are similar. Since $BC || DE$ and $BF$ is an angle bisector of $\angle \, A$ $$\angle \,DBF = \angle \, CBF = \angle \, DFB$$ so triangles $BDF$ is isosceles with $BD = DF$. Analogously $CE=EF$. Hence the perimeter $P_{ADE}$ of triangle $ADE$ is $$P_{ADE} = AD+DF+AE+EF = AD+DB + AE+EC = AB + AC = 26 + 34 = 60$$ The perimeter $P_{ABC}$ of $ABC$ is $$P_{ABC} = AB + BC+AC = 26+4=+54 = 100$$ By the similarity of $ADE$ and $ABC$ $$\frac{DE}{BC} = \frac{P_{ADE}}{P_{ABC}} = \frac{60}{100} = \frac{3}{5}$$ Since $BC = 40$ $$DE = \frac{3}{5} \, 40 = 24$$

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    nice use of the parallel lines – 2017-02-24
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    Brilliant use of basic geometry! – 2017-02-24
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    @Jim thank you! I appreciate it. Cheers! – 2017-02-25
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You can do this with the angle bisector theorem used twice.

First observe that $AF$ bisects $\angle BAC$, (because angle bisectors are concurrent) so continue $AF$ to meet $BC$ at point $G$. Then $G$ divides $BC$ in the ratio 26:34 so $BG = \frac {40}{60} 26 =\frac {52}{3}$

Then $BF$ divides $AG$ in the ratio $26:\frac{52}3 = 3:2$ giving $AF:AG$ as $3:5$. Thus through similarity of $\triangle ABC$ and $\triangle ADE$ the ratio between $DE$ and $BC$ is also $3:5$ i.e. $\fbox{$DE=24$}$

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    Can you explain what you mean by bisectors are concurrent? Also, how you know that it divides BC in a ratio of 26:34? Thank you! – 2017-02-24
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    The angles bisectors of a triangle [meet at a common point](https://www.algebra.com/algebra/homework/Triangles/Angle-bisectors-of-a-triangle-are-concurrent.lesson) ("are concurrent"). The [angle bisector theorem](https://en.wikipedia.org/wiki/Angle_bisector_theorem) says that the bisector divides the opposite side in the ratio of the adjacent sides. – 2017-02-24
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    @NickBrown I proved [the angle bisector theorem here](http://math.stackexchange.com/questions/2111870/what-is-the-formula-for-bisector-of-triangle/2111929#2111929) and [the concurrency of angle bisectors here](http://math.stackexchange.com/questions/2106376/bisectors-of-a-triangle-meet-at-point/2106387#2106387) – 2017-02-24
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    Ok, so I follow everything until the last part where you find DE and BC to be in a 3:5 ratio. Where did you get that from? – 2017-02-24
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Try with Heron's formula to get the area of the big triangle two ways:

  • First calculate it as a function of the three sides, using Heron's directly.

  • Calculate the height of the triangle ABC $=h_{ABC}$ from the areas just calculated.

  • Then calculate the radius of the incircle, using this example and knowledge of the relationship of the incircle to the bisected angles of the triangle.

    • subtract this radius from the height of ABC $=h_{ABC}$ to get the height of triangle ADE $=h_{ADE}$.

  • Now use proportionality of similar triangles:

$$\frac{h_{ADE}}{h_{ABC}}= \frac{|DE|}{40}$$

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    Can you explain why subtracting the inradius from the height of ABC would give me the height of triangle ADE? – 2017-02-24
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    The in radius is the height of triangle BFC. The height of ADE is the height of ABS minus the height of BFC, since the line DE is parallel to the line BC. – 2017-02-24
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Let $S$ be the area of the triangle, let $p$ be its perimeter, $r$ its inradius, and $h_A$ its altitude from $A$. Also write $BC = a$, $AC = b$, $AB = c$.

Note that $F$ is the incentre of $ABC$, hence the distance from $F$ to $BC$ is $r$. Now we have $$\frac{DE}{a} = 1 - \frac{r}{h_A} = 1 - \frac{2S/p}{2S/a} = 1 - \frac{a}{p} = \frac{b + c}{p}$$ hence $$DE = \frac{a(b+c)}{p} = \frac{40(34 + 26)}{100} = 24.$$