[This is a geometry puzzle, solvable by elementary 8th grade geometry, i.e. no trigonometry etc]
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]
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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