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Suppose there is an isosceles triangle ABC with $\angle$B$=20$ degrees, $AB=BC$

M,N are points on AB,BC such that $\angle$MCA=60 degrees, $\angle$ NAC=50 degrees. Find $\angle$NMC

How do you even start? Many angles can be found by using angle sum property of the triangle but I can't get anywhere beyond that. I tried constructing a parallel through both M and N but it doesn't help.

Also, one could repeatedly use sin rule to find it out, but this is extremely inefficient.

Please give me a hint...

aman
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  • I'm so sorry. $AB=BC$ – aman Sep 02 '19 at 11:00
  • This is a fun problem to solve, but it is not an angle-chasing problem. You can Google "world's hardest easy geometry problem" for analysis on this and it's sister problem where the internal angles are 70 and 60. –  Sep 02 '19 at 11:16

1 Answers1

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HINT

The lines $MC$ and $NA$ cross at a point I'll call $X$ in the middle of the triangle. The "vertical angle" theorem at $X$ tells you, for instance, that $MXN$ and $AXC$ are equal angles.

Can you determine the angles $BAC$ and $BCA$? Hint: isoceles.

If you draw the line $MN$, then at the point $M$, there are three angles ($BMN, NMX, XMA$) that sum to $180$ degrees, and similarly at $N$.

From those facts along, you should be able to determine the answer.

John Hughes
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  • I found $BAC$ and $BCA$ instantly: They're both 80 degrees. So, $AXC=MXN=70$ Now, We can find all the angles in triangles $NCM$ and $MXA$ Then what? In order to find the rest, what do we do? – aman Sep 02 '19 at 10:59
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    Ah... I see. Well, it's probably kinder to say what things you've computed already rather than having us work through step by step suggestions only to have you say "been there, done that" in response. Regardless, @MatthewDaly's comment clears this up. – John Hughes Sep 02 '19 at 11:21