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Philippine Mathematical Olympiad Area Stage

Geometry Difficulty 4.9 AIME Prove it Philippines

Problem:

Points $A$ , $M$ , $N$ and $B$ are collinear, in that order, and $AM = 4$ , $MN = 2$ , $NB = 3$ . If point $C$ is not collinear with these four points, and $AC = 6$ , prove that $CN$ bisects $\angle BCM$ .

Solution

Solution:

Since $\frac{CA}{AM} = \frac{3}{2} = \frac{BA}{AC}$ and $\angle CAM = \angle BAC$ , then $\triangle CAM \sim \triangle BAC$ . Therefore,
$\angle MCA = \angle CBA.$
Since $AC = 6 = AN$ , then $\triangle CAN$ is isosceles. Therefore,
$\angle ACN = \angle ANC.$
Thus,
$\begin{array}{rlr} \angle BCN & = \angle ANC - \angle CBA \quad \text{since } \angle ANC \text{ is an exterior angle of } \triangle BNC \\ & = \angle ACN - \angle MCA \quad \text{using (1) and (2)} \\ & = \angle MCN. & \end{array}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.