In $\triangle ABC, AB = 28, BC = 21$ and $CA = 14$. Points $D$ and $E$ are on $AB$ with $AD = 7$ and $\angle ACD = \angle BCE$

Question

In $\triangle ABC, AB = 28, BC = 21$ and $CA = 14$. Points $D$ and $E$ are on $AB$ with $AD = 7$ and $\angle ACD = \angle BCE$. Find $BE$.

What I Tried: Here is a picture :-

I know the side-lengths of the triangle so I can find out their altitudes too using Heron's Formula, but that didn't give me any useful information. The fact is that I can't use the side-lengths in any way, neither the triangles, because there are no similar triangles here. Angle-chasing, I don't think it is going to help. I did not try Trigonometry cause I am a little weak at it.

Can anyone give me any ideas for this problem? Thank You.

Answer 1

We have $\triangle DAC \sim \triangle CAB$ by SAS ($\angle A$ common)

$$\therefore \angle B = \angle ACD = \angle BCE$$

So $\triangle BEC$ is isosceles with $BE=CE$. Drop $EF \perp BC$.

$\triangle BEF \cong \triangle CEF$

$BF = 21/2$

By cosine rule in $\triangle ABC$,

$$ \cos B = \dfrac{28^2 + 21^2 - 14^2 }{2\cdot 28 \cdot 21} = \dfrac{7}{8} $$

Easy enough,

in right $\triangle BEF$, $$ BE \cos B = BF $$ $$ \Rightarrow \boxed{BE = 12}$$