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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
3 years, 5 months ago

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Just do the angle chasing. Very easy.

Let \(B\hat{A}C = x\).

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

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

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

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

\(\angle BAC : \angle NBA = 2 : 3\)

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

\(x = 20\) ~~~

Samuraiwarm Tsunayoshi - 3 years, 5 months ago

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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).

Mathh Mathh - 3 years, 5 months ago

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Please provide a full solution :D

Victor Loh - 3 years, 5 months ago

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