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I am having trouble finding a proper solution to this problem:

An equilateral triangle (ABC) is inscribed in a circle (o). Point D is in the shorter BC arc of circle o. Point E is symmetric to point B about line CD. Prove that points A, D, E are in the same straight line.

What I already tried: In short - I noticed a rhombus and proved that ADE are in the same straight line, because the angle $$ \measuredangle{ADE}=180^o $$ My solution is wrong.

Regards, Tom.

Harsh Kumar
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TomDavies92
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3 Answers3

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Let $F$ be the point where $CD$ intersects $BE$. This proof relies on inscribed angles.

We have $\angle CDB = 120^\circ$. That means $\angle BDF = 60^\circ$. Therefore, we also have $\angle FDE = 60^\circ$ by symmetry. So $\angle BDE = 120^\circ$. Also, we have $\angle ADB = \angle ACB = 60^\circ$. Therefore we arrive at $\angle ADE = 180^\circ$.

Arthur
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Hint:

  • When you reflect $B$, you get that $|\angle CDE|$ = $|\angle CDB|$.
  • Angles $\angle CDB$ and $\angle CDA$ are based on very specific arcs.

I hope this helps $\ddot\smile$

dtldarek
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Well, ∠CBA = ∠ACB = 60 deg. => arc AC = arc AB = 120, thus the big arc CB = 240, that means ∠CDB = 1/2 arc CB = 120, because it is an inscribed angle.

Martin Shishkov
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