How can I find the sum of the angle $AMB$, angle $ANB$ and the angle $ACB$?

Question

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 $AB=BM$.

Here are 2 of my attempts

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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

Answer 1

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 $$\alpha+\beta+\gamma=90^{\circ}.$$

Answer 2

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 $90^\circ$.