The 80-80-20 Triangle Problem, A Derivative, Solution #10

$ABC\;$ is an isosceles triangle with vertex angle $\angle BAC = 20^{\circ}\;$ and $AB = AC.\;$ Point $E\;$ is on $AB\;$ such that $AE = BC.\;$ Find the measure of $\angle AEC.$

Solution

Construct an equilateral $\Delta ADE$ outside $\Delta ABC.$

Since $\angle CAD=80^{\circ},\;$ $AD=AE=BC,\;$ $\Delta ACD=\Delta ABC.$ In particular, $\angle CED=\angle AEC.$ Thus $2\angle AEC+60^{\circ}=360^{\circ}.$ It follows that $\angle AEC=150^{\circ}.$

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