# Prove that $\angle{CBM}=60^{\circ}-\frac{n}{2}$

Given a $\triangle{ABC}$ with $\angle{BAC}=2\cdot \angle{ACB}=n^{\circ}$ where $0<n<120$, let $M$ be an interior point of $\triangle{ABC}$ with $BA=BM$ and $MA=MC$. Prove that $\angle{CBM}=60^{\circ}-\frac{n}{2}$.

Attempt:

We know that $\alpha+\beta = \angle{BAC} = n^{\circ}$ and that $$\angle{BAC}+\angle{ACB}+\angle{CBA} =\dfrac{3}{2}(\alpha+\beta)+\angle{CBA} = 180^{\circ}.$$ Also, $\angle{CBM} = \angle{CBA}-180^{\circ}+2\alpha$ and thus $$\dfrac{3}{2}(\alpha+\beta)+\angle{CBA} = \dfrac{3}{2}(\alpha+\beta)+\angle{CBM}+180^{\circ}-2\alpha =180^{\circ}$$ which means that $$\dfrac{3}{2}(\alpha+\beta)+\angle{CBM} = 2\alpha$$ and $$\angle{CBM} = \dfrac{1}{2}(\alpha-3\beta) = \dfrac{1}{2}(n-4\beta).$$ How do you continue from here?

Draw the line $BD$ parallel to $AC$ and choose $D$ such that $CD=BD$, then try to prove that $BMD$ is an equilateral triangle, then the solution will follow.

Proof

From the construction we have that $\angle BCD=\angle DBC= n/2$, hence $$\angle ACD=n/2+n/2=n= \angle BAC.$$

It follows that $ABDC$ is an isosceles trapeziod, then $AB=CD$. This implies that $ABM$ and $CMD$ are congruent triangles, in particular $$CD=MD=MB=AB.$$ Fnally $BD=CD$ by construction, so $$CD=MD=MB=AB=BD.$$

• How does the solution follow? – user19405892 May 5 '16 at 21:50
• Because then $\angle CBM=\angle MBD - \angle CBD=60 - n/2$. – mrprottolo May 5 '16 at 21:54
• Why does $\angle{CBD} = \dfrac{n}{2}$? – user19405892 May 5 '16 at 21:54
• Because it is equal to $\angle ACB$ since $BD$ is parallel to $AC$. – mrprottolo May 5 '16 at 21:56
• Can you tell me how to prove it is equilateral? – user19405892 May 5 '16 at 22:11