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Let $ABC$ be a triangle with circumcicle $\omega$. Then the bisector of $\angle ABC$ meets $AC$ at $D$ and circle $\omega$ at $M \ne B$. The circumcircle of $\triangle BDC$ meets $AB$ at $E\ne B$, and $CE$ meets $\omega$ at $P\ne C$. The bisector of $\angle PMC$ meets $AC$ at $Q \ne C$. If $PQ = MC$, what is $\angle ABC$?

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I've removed my comment. For the future: If you make a change that makes someone else's answer or comment look wrong, please mark it as such. –  joriki Sep 25 '12 at 22:33
    
Forgive my ignorance, but isn't $Q\neq C$ obvious? For the bisector of any angle to intersect one of its endpoints that isn't its vertex, the angle must be 0. –  KeithS Sep 25 '12 at 23:48

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This sounds like homework. Here's a hint: use the inscribed angle theorem, which states two things of interest to you:

  • The angle $\angle ABC$ inscribed within a circle by points $A$, $B$, and $C$ located on the edge of the circle is always exactly half of the central angle of the intercepted arc between B and C.
  • Opposing vertices of any quadrangle that can be inscribed by a circumcircle (a cyclic quadrangle) have supplementary angles.

These allow you to form the following lemmas:

  • The arcs $AM$ and $MC$ each have the same central angle measure as $\angle ABC$.
  • The angles $\angle EDC$ and $\angle EBC$ are supplementary, and the angles $\angle BED$ and $\angle BCD$ are supplementary, because by definition they form a quadrangle within the circumcircle of $\triangle BCD$.
  • By the same token, angles $\angle PBC$ and $\angle PMC$ are supplementary, and $\angle BPM$ and $\angle BCM$ are supplementary.

Here's something that looks right to me, but my geometry is insufficient to prove it; I think if you can, it'll go a long way:

  • Quadrangles $BCDE$ and $PMCB$ are similar; $BCDE$ is upside-down relative to $PMCB$. Prove this, and you get $\angle ABC \cong \angle PMC$ and $\triangle PMC \cong \triangle ABC$, therefore $MC = BC$.
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