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# An interesting Geometry problem

Let $$ABC$$ be a triangle such that $$AB=AC$$. Let $$M$$ be a point on $$AB$$ such that $$MA=MC$$ and let $$N$$ be a point on $$AC$$ such that $$CN=CB$$ and $$\angle BAC:\angle NBA=2:3$$. Calculate the size of $$\angle NMC$$.

Note by Victor Loh
2 years, 8 months ago

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Just do the angle chasing. Very easy.

Let $$B\hat{A}C = x$$.

$$\displaystyle \overline{AB} = \overline{AC}; A\hat{B}C = A\hat{C}B = 90 - \frac{x}{2}$$

$$\displaystyle \overline{MA} = \overline{MC}; M\hat{C}A = x$$

$$\displaystyle \overline{CN} = \overline{CB}; C\hat{N}B = C\hat{B}N = 45 + \frac{x}{4}$$

$$\displaystyle \angle ABC; N\hat{B}A = 45 - \frac{3x}{4}$$

$$\angle BAC : \angle NBA = 2 : 3$$

$$\displaystyle 3x = 2(45 - \frac{3x}{4})$$

$$x = 20$$ ~~~ · 2 years, 8 months ago

You've found $$\angle BAC = 20^{\circ}$$, but the problem is asking for $$\angle NMC$$. Actually, after your solution this problem becomes literally equivalent to the one presented in Moscow 1952 and even has a name Langley's Adventitious Angles (an animated explanation here). · 2 years, 8 months ago