8
$\begingroup$

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.

$\endgroup$
8
$\begingroup$

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 $\angle ADE = 180^\circ$.

$\endgroup$
0
5
$\begingroup$

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 $\ddot\smile$

$\endgroup$
1
  • 2
    $\begingroup$ That is a cool smiley face. $\endgroup$ – Joao Dec 2 '14 at 2:11
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.