## The 80-80-20 Triangle Problem, 40-50 Variant, Solution #1

Let ABC be an isosceles triangle |

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Copyright © 1996-2017 Alexander Bogomolny

This is one modification of the original problem that admits a practically trivial solution.

Since ∠BEC = 180° - 80° - 50° = 50°, ΔBCE is isosceles. In addition,

Thus BD divides CE in two and is perpendicular to it. Applying this information to ΔCDE we see that DB is the the altitude and also the median from D meaning that ΔCDE is isosceles:

∠CED = ∠DCE = 80° - 50° = 30°. |

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Copyright © 1996-2017 Alexander Bogomolny

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