In $\Delta ABC$ $\angle BAC = 40^\circ, AB = 10$ and $AC = 6$. Points $D$ and $E$ lie on $AB$ and $AC$ respectively.

Question

In $\Delta ABC$, $\angle BAC = 40^\circ, AB = 10$ and $AC = 6$. Points $D$ and $E$ lie on $AB$ and $AC$ respectively. What is the minimum possible value of $BE + DE + CD?$

What I Tried: Here is a picture :-

I have already seen the solution in the Art of Problem Solving Website, because this is one of the $2014$ AMC $12$A Problems. Check the solutions here :- https://artofproblemsolving.com/wiki/index.php/2014_AMC_12A_Problems/Problem_20

The problem, however, is that all the solutions use Law of Cosines as Trigonometry, and I prefer using simple geometry techniques to solve this problem.

So is it possible to solve this without Trigonometry? Thank You.

Answer 1

In this construction we have $|AB_1|=|AB|=10$, $|AC_1|=|AC|=6$, $|ED_1|=|ED|$, $|D_1C_1|=|DC|$.

Since $\angle C_1GA=90^\circ$ and $\angle GAC_1=180^\circ-3\cdot40^\circ=60^\circ$, $\triangle C_1GA$ is half of the equilateral triangle with the altitude $C_1G$, hence \begin{align} |AG|&=\tfrac12\,|AC_1|=3 ,\\ |C_1G|&=\sqrt{|AG|^2+|AC_1|^2}=\tfrac{\sqrt3}2\,|AC_1|=3\sqrt3 ,\\ |BG|&=|AG|+|AB|=13 ,\\ |BC_1|&=|BE|+|ED_1|+|D_1C_1| \\ &=|BE|+|ED|+|DC| \\ &=\sqrt{|BG|^2+|C_1G|^2}=\sqrt{196}=14 . \end{align}