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

Question

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?

mrprottolo · Accepted Answer · 2016-05-05T22:26:27.110

5

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 $$CD=MD=MB=AB=BD.$$

edited May 05 '16 at 22:26

answered May 05 '16 at 21:22

mrprottolo

3,792

How does the solution follow? – user19405892 May 05 '16 at 21:50
Because then $\angle CBM=\angle MBD - \angle CBD=60 - n/2$. – mrprottolo May 05 '16 at 21:54
Why does $\angle{CBD} = \dfrac{n}{2}$? – user19405892 May 05 '16 at 21:54
Because it is equal to $\angle ACB$ since $BD$ is parallel to $AC$. – mrprottolo May 05 '16 at 21:56
Can you tell me how to prove it is equilateral? – user19405892 May 05 '16 at 22:11
@user19405892 I added a short proof. – mrprottolo May 05 '16 at 22:26
How did you conclude that $\triangle{ABM} \cong \triangle{CDM}$? What angle did you use? – user19405892 May 05 '16 at 22:45
I used $\angle DCM=\angle MAB$. – mrprottolo May 05 '16 at 22:47

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

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