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How can I find the sum of the $\angle AMB, \angle ANB$ and the $\angle ACB$? In triangle $ABC$, $\angle ABC =90^\circ$. $BC$ is divided in $3$ parts such that $BM=BN=NC$. And also $AB=BM$.

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Here are 2 of my attempts

prikachi.com/images.php?images/182/9546182c.jpg

prikachi.com/images.php?images/183/9546183U.jpg

But kinda messed up

I found their sum here prikachi.com/images.php?images/601/9546601o.jpg

But I really want to understand the Michael's method

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closed as off-topic by Saad, Javi, Paul Frost, Adrian Keister, InterstellarProbe Apr 15 at 18:52

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Let $\Delta CMK\cong \Delta NBA$ such that $A$ and $K$ placed in the different sides respect to $BC$.

Draw it in the checkered page!

Thus, $AK=CK$, $$\measuredangle CKA=\beta+90^{\circ}-\beta=90^{\circ},$$ $$\gamma+\beta=\measuredangle ACK=45^{\circ}$$ and $$\alpha+\beta+\gamma=90^{\circ}.$$

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  • $\begingroup$ Ok I understand but how prove that ΔCMK≅ΔNBA $\endgroup$ – rucan1 Apr 16 at 6:47
  • $\begingroup$ I found their sum here prikachi.com/images.php?images/601/9546601o.jpg But I really want to understand your method $\endgroup$ – rucan1 Apr 16 at 6:49
  • $\begingroup$ @rucan1 Draw it in the checkered page! Нарисуйте этот треугольник на листочке в клеточку. שרטט את המשולש בדף משובץ $\endgroup$ – Michael Rozenberg Apr 16 at 6:58
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Note that the angle $AMB=45^\circ$ because $AB=BM$.

For the rest, we see that $ANB=\tan^{-1}\frac{1}{2}$ and $ACB=\tan^{-1}\frac{1}{3}$. Thus: $$ANB+ACB=\tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3}=\tan^{-1}\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=\tan^{-1}1=45^\circ.$$ Therefore the desired sum is $90^\circ$.

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  • $\begingroup$ Thank you! And Also is there an easier way to be solved/explained for high school students. $\endgroup$ – rucan1 Apr 15 at 10:25

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