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
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