The 80-80-20 Triangle Problem, A Derivative, Solution #4
ABC is an isosceles triangle with vertex angle
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Copyright © 1996-2018 Alexander Bogomolny
This is a trigonometric solution reported in [Leikin].
Let E' be a point on AB with
By the Law of Sines, in ΔACE',
AE' / sin 10° = CE' / sin 20°.
But, by the double angle formula,
CE' = 2·AE'·cos 10°.
In ΔBCE',
BC / sin 30° = CE' / sin 80° = CE' / cos 10°,
so that
CE' = 2·BC·cos 10°,
implying AE' = BC and, subsequently, E = E'. Thus
Reference
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Copyright © 1996-2018 Alexander Bogomolny
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