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 ∠BCM.
I already solved this using Stewart's Theorem, but I tried to solve it using Sine Law but I couldn't. This made me think that whenever I apply sine law I couldn't do it fast/efficiently
I would want to show my solution, but its all random triangles and angles that I used with no basis, I just couldn't find a way with all of my expressions. I did notice that sin(a+b) = sin (c+d), and that we can use the fact that $sin(a)/sin (b+c+d) = 4/6 which is also the ratio of x and y if b and c are equal angles.
Note that you have an isosceles triangle $ACN$ since $AC=AN=6$. So, you have that $a+b=c+d$. You want to prove that $b=c$. So it is enough to prove that $d=a$. This will be true since $6^2=4(4+5)$.
Let me explain: Law of sines in $AMC$ gives $\sin(\pi-e-a)/\sin a=6/4=3/2$. And Law of sines in $ACN$ gives $\sin(\pi-e-d)/\sin d=9/6=3/2$.
Deduce from here that $a=d$