A geometry problem involving triangles

Question

In the figure, AE is the bisector of the exterior angle CAD meeting BC produced in E. If AB = 10 cm, AC = 6 cm and BC = 12 cm, then find the length of CE.

My Attempt: I tried to find out the existence of congruent triangles in the diagram, but couldn't find any.

Any help will be appreciated.

Definitely not. (Evident from the given lengths of AB, BC and CA) — MathsLearner, Jul 01 '17 at 00:22
from cosine theorem you can find all angles from triangle $ABC$ and solve your OP — serg_1, Jul 01 '17 at 00:23

Huang · Answer 1 · 2017-07-01T00:54:42.040

Hint : The cosine law of triangles will definitely solve the problem but that may lead to boring calculations.

You can construct congruent triangles.

Draw a line passing point E and parallel to AC, intersecting AB at F. Now ABC is congruent to FBE, and AF=EF, so you can find the length of AF by solving the equation w.r.t AF.

$\frac{BA}{BA+AF}=\frac{AC}{EF}=\frac{AC}{AF}$

Continue the reasoning and you can find the length of CE.

score 1 · Answer 2 · answered Jul 01 '17 at 03:12

1

Let $CE=x$.

Hence, by bisector theorem we obtain: $$\frac{AB}{AC}=\frac{BE}{EC}$$ or $$\frac{10}{6}=\frac{x+12}{x}$$ or $$\frac{5}{3}=1+\frac{12}{x}$$ or $$\frac{12}{x}=\frac{2}{3}$$ or $$x=18$$.

answered Jul 01 '17 at 03:12

Michael Rozenberg

194,933

score 1 · Accepted Answer · answered Jul 01 '17 at 04:10

Alternatively: Continue the line $D$ until $F$ and connect $F$ with $E$ so that $AC=AD$.

Note that $\Delta ACE$ is equal to $\Delta ADE$, because two sides and the angle between them are equal. It implies the line $AE$ is a bisector in $\Delta BDE$.

Using the property of bisector: $$\frac{AB}{BE}=\frac{AD}{DE} \Rightarrow \frac{10}{12+CE}=\frac{6}{CE} \Rightarrow CE=18.$$

A geometry problem involving triangles

3 Answers3