Length of segment parallel to an edge 
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.
 A: 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 $BC = 40$
$$DE = \frac{3}{5} \, 40 = 24$$
A: 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$}$
A: 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:
$$\frac{h_{ADE}}{h_{ABC}}= \frac{|DE|}{40}$$
A: 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 $$DE = \frac{a(b+c)}{p} = \frac{40(34 + 26)}{100} = 24.$$
