My Math Forum Prove that two angles are congruent

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May 16th, 2018, 10:16 AM   #1
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Prove that two angles are congruent

Quote:
 Circles $\displaystyle k_{1}$ and $\displaystyle k_{2}$ intersect at points $\displaystyle A$ and $\displaystyle B$. Straight line $\displaystyle p$ that contains point $\displaystyle B$ intersects circles $\displaystyle k_{1}$ and $\displaystyle k_{2}$ at two more points $\displaystyle M$ and $\displaystyle K$, respectively. Tangent line to $\displaystyle k_{1}$ in point $\displaystyle M$ and tangent line to $\displaystyle k_{2}$ at point $\displaystyle K$ intersect at point $\displaystyle C$. Prove that $\displaystyle \measuredangle MAC=\measuredangle BAK$.
I managed to prove the following.
$\displaystyle \measuredangle MBA=180^{\circ}-\measuredangle ABK=\measuredangle AFK$ (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.
Attached Images
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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 $\angle MCA = \angle MKA$. Thanks from lua

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