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Let $\triangle ABC$ be an isosceles triangle ($AB = AC$ and $\angle ABC = \angle ACB = 35^\circ$). We have a point $M$ inside the triangle such that $\angle MBC = 30^\circ$ and $\angle MCB = 25^\circ$. Find the measure of $\angle AMC$.

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closed as off-topic by Claude Leibovici, user91500, gebruiker, John B, Daniel W. Farlow May 16 '16 at 12:38

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Let $S$ be the circumcenter of triangle $BCM$. Then $\angle CSM=2\angle CBM=60^\circ$. Therefore triangle $SCM$ is equilateral. In particular $CM = SC$.

Observe now that $\triangle BCS$ is isosceles with $BS=CS$ and $\angle BCS = 60^\circ - 25^\circ = 35^\circ$. Therefore triangles $BCS, BCA$ are congruent. In particular $SC=AC$.

Therefore $CM=AC$. Looking at the triangle $ACM$ we find $$\angle CMA = \frac{180^\circ - \angle ACM}{2} = \frac{180^\circ - 10^\circ}{2} = 85^\circ.$$

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  • $\begingroup$ Typo: triangle BCS is isosceles, not triangle BCM. $\endgroup$ – almagest May 15 '16 at 11:55
  • $\begingroup$ Thank you for careful proofreading. I have corrected the typo. $\endgroup$ – timon92 May 15 '16 at 11:57
  • $\begingroup$ Thank you very much. I would appreciate it if you answer my other question: math.stackexchange.com/questions/1786050/… $\endgroup$ – titansarus May 15 '16 at 13:08

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