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: enter image description here

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?


enter image description 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.


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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.