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.

• Correct answer is 24, I also came to 40 – Nick Brown Feb 24 '17 at 1:45
• I removed my comment because I realised how stupid it was to suggest that DE = BC lol - but I see the mistake I made – mrnovice Feb 24 '17 at 1:46

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 $BC = 40$ $$DE = \frac{3}{5} \, 40 = 24$$

• nice use of the parallel lines – Joffan Feb 24 '17 at 5:05
• Brilliant use of basic geometry! – Jim Feb 24 '17 at 17:38
• @Jim thank you! I appreciate it. Cheers! – Futurologist Feb 25 '17 at 11:18

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$}$

• 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! – Nick Brown Feb 24 '17 at 1:52
• The angles bisectors of a triangle meet at a common point ("are concurrent"). The angle bisector theorem says that the bisector divides the opposite side in the ratio of the adjacent sides. – Joffan Feb 24 '17 at 2:03
• @NickBrown I proved the angle bisector theorem here and the concurrency of angle bisectors here – Joffan Feb 24 '17 at 2:10
• 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? – Nick Brown Feb 24 '17 at 2:38

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:

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

• Can you explain why subtracting the inradius from the height of ABC would give me the height of triangle ADE? – Nick Brown Feb 24 '17 at 2:30
• 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. – Jim Feb 24 '17 at 5:54

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 $$DE = \frac{a(b+c)}{p} = \frac{40(34 + 26)}{100} = 24.$$