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May 16th, 2018, 10:16 AM   #1
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Joined: Jan 2018
From: Belgrade

Posts: 55
Thanks: 2

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 Drz2015_VII_z2.jpg (32.1 KB, 12 views)

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 Tags angles, congruent, prove Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post greenleaf Geometry 3 September 27th, 2015 06:00 AM mathfixxer Geometry 5 March 24th, 2015 05:55 PM claudiohorvi Number Theory 5 November 25th, 2013 06:15 AM coffee_leaf Number Theory 1 September 2nd, 2012 09:07 PM smslca Number Theory 4 January 28th, 2012 10:21 PM

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