Right triangle $ABC$ has $AC = 8$ and $CB = 6$. $M$ is the midpoint of $AB$. Pick point $N$ on line $CM$ with $M$ between $C$ and $N$ such that $∠CAB = ∠BAN$. Compute $MN$. Express your answer as a common fraction.

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I figured out that $CM$, $BM$, and $AM$ were $5$, but I can't figure out how to continue.


I think it is better to leave the trigonometry out of the solution.

Since $\angle BCA = 90^{\circ}$ one can conclude that $AM = BM = CM$. By Pythagoras one gets $AB = 10$ so $AM = BM = CM = 5$. Triangle $ACM$ is isosceles and $$\angle \, ACN = \angle \, ACM = \angle\, CAM = \angle \, CAB = \angle \, BAN = \angle \, MAN$$ Let's look at triangles $ACN$ and $AMN$. We have concluded that $\angle\, ACN = \angle \, MAN$. Since $\angle \, ANM = \angle \, CNA$ as a common angle at vertex $N$, triangles $ACN$ and $AMN$ are similar and therefore $$\frac{AM}{CA} = \frac{MN}{AN} = \frac{AN}{CN}$$ However, since $AM = 5, \,\,$ $AC=8$ and $CN = CM + MN = 5 + MN$, we get the relations $$\frac{5}{8} = \frac{MN}{AN} = \frac{AN}{5+MN}$$ Thus, one can express $AN$ in two different ways $$AN = \frac{8}{5}\, MN \,\,\, \text{ and } \,\,\, AN = \frac{5}{8}\, (MN + 5)$$ Equate the two, obtaining the equation $$\frac{8}{5}\, MN = \frac{5}{8} (MN + 5)$$ Solve the equation for $MN$ in order to obtain $$MN = \frac{125}{39}= 3 + \frac{8}{39}$$


We know $CM = 5$.

Also let $\angle CAM = a$. Then we have

  • $\angle MAN=a$,
  • $\angle AMN = 2a$ and
  • $\angle MNA = 180-3a$

Now use the sine rule in $\triangle AMN $ and $\triangle AMC$

$$\dfrac{\sin(180-3a)}{AM} = \dfrac{\sin(a)}{MN}$$

and $$\dfrac{\sin(a)}{CM} = \dfrac{\sin(180-2a)}{AC}$$

  • $\begingroup$ Okay, so that leaves me with sin(a)/5=sin(180−2a)/8, and sin(180-3a)/5=sin(a)/MN, but how can I use that information? Btw sorry for my inability to code and make nice formatting – I'm new to the community $\endgroup$ – Wiifan Mar 31 '17 at 7:35
  • $\begingroup$ @Wilfan Those are two equations in two unknowns. Find $\sin(a)$ from one, then you have $\sin(2a)$ and $\sin(3a)$ using trigonometry. Then plug those values to find $MN$. For basic LaTeX guide you can refer here. $\endgroup$ – samjoe Mar 31 '17 at 7:43
  • $\begingroup$ Oh ok. Thank you for your help and for showing me the ropes! $\endgroup$ – Wiifan Mar 31 '17 at 7:59

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