Solution to the two triangles puzzle.

[This is a solution to the two triangles geometry puzzle posted earlier]

The problem, repeated here:

In the figure below (not to scale, forgive the shoddy drawing skills).




$ABC$ is a triangle such that $\angle{BAC} = 60$ and $\angle{ABC} = 25$.

$DEF$ is an isosceles triangle, such that $\angle{EDF} = \angle{EFD}$, and $\angle{DEF} = 10$

(all angles are in degrees).

We also have that $|BC| = |DE|$ ($|XY|$ = length of the segment $XY$)

Show that $2|AC| + |DF| = |AB|$

Solution

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Notice that $\angle{FDE} = 85$ and $\angle{ACB} = 95$, and so their sum is $180$.

Since $|BC| = |ED|$, we can position one copy of $\triangle{ABC}$ with $B$ coinciding with $E$ and $C$ coinciding with $D$ to get a bigger triangle.

Another copy of $\triangle{ABC}$, call it $\triangle{A'B'C'}$, can be positioned so that $B'$ coincides with $E$ and $C'$ coincides with $F$.

The result will be an even bigger triangle, $\triangle{EAA'}$ as below (more drawing incompetence):

Placing two copies of ABC along with DEF results in an equilateral triangle.

$\triangle{EAA'}$ is an equilateral triangle, and thus the base of the triangle which is $2|AC| + |DF|$ is same as the other side which is $|AB|$.