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 $\angle NMC$.

#Geometry

Easy Math Editor

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Comments

Please provide a full solution :D

Victor Loh - 6 years, 10 months ago

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 - 6 years, 10 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).

mathh mathh - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$