MOG 2015 2

2

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First note that AB=BE=BC as the nonagons are regular so \triangle ABC is isosceles.

\begin{align} \angle ABC &= 2\times \text{external angle of regular nonagon}\\ &= 2\times \frac{360^\circ}9\\ &=80^\circ\\\\ \end{align}

Since \triangle ABC is isosceles:

\begin{align} \angle BCA &= \frac{180^\circ-80^\circ}2\\ &=50^\circ \end{align}

Considering angles about C:

\begin{align} \angle ACD &= 360^\circ-50^\circ-\text{internal angle of nonagon}-\text{internal angle of hexagon}\\ &=360^\circ-50^\circ-140^\circ-120^\circ\\ &=50^\circ \end{align}

Now, since BC=CF=CF, \angle BCA=\angle ACD and AC is shared, the two triangles are congruent and therefore both isosceles.

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