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May 16th, 2018, 10:16 AM  #1  
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2  Prove that two angles are congruent Quote:
$\displaystyle \measuredangle MBA=180^{\circ}\measuredangle ABK=\measuredangle AFK$ (peripheral angle over AK). This means that peripheral angles over arcs AM and AK are congruent, which means that coresponding central angles are congruent, too: $\displaystyle \measuredangle MO_{1}A=\measuredangle AO_{2}K$ Now, $\displaystyle \measuredangle AMO_{1}=\measuredangle MAO_{1}=\measuredangle KAO_{2}=\measuredangle AKO_{2}$ (that is, $\displaystyle \triangle MAO_{1} \sim \triangle KAO_{2}$). On the other hand, $\displaystyle \measuredangle AMC=\measuredangle AMB+\measuredangle BMC=\measuredangle AMB+\measuredangle MAB=180^{\circ}\measuredangle MBA=\measuredangle ABK$. I still need to prove that $\displaystyle \measuredangle MCA=\measuredangle BKA$ to finish the proof, but don't see how. Last edited by lua; May 16th, 2018 at 10:34 AM.  
May 19th, 2018, 09:59 PM  #2 
Member Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 
Anybody any idea?

May 20th, 2018, 01:31 PM  #3 
Global Moderator Joined: Dec 2006 Posts: 20,921 Thanks: 2203 
As $\angle ABK = 180^\circ  \angle AFK = 180^\circ  \angle AKC$, you can conclude that $AKCM$ is a cyclic quadrilateral. Hence $\angle MCA = \angle MKA$. 

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angles, congruent, prove 
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