# How can I find the sum of the angle $AMB$, angle $ANB$ and the angle $ACB$? [closed]

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

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

## closed as off-topic by Saad, Javi, Paul Frost, Adrian Keister, InterstellarProbeApr 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 $$\alpha+\beta+\gamma=90^{\circ}.$$

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

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 $$90^\circ$$.

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