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I have to solve the following problem:

Consider an equilateral triangle $ABC$ and $\mathcal{C}$ its circumscribed circle. Let $M$ be a point located on the arc of the circle defined by $[AC]$ which does not contain $B$.
1) Show that $BM$ is the bisector of the angle $\widehat{AMC}$.
2) Let $D$ be a point of $[BM]$ such that $DM=DA$. Show that the triangle $DAM$ is equilateral.
3) Show: $MA + MC = MB$.

I was able to prove 1) and 2), but I am stuck at 3). Can anyone help me with point 3). Thank you in advance!

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we will use that an arc makes same angle on the circumference. $\angle BMC = \angle BAC = 60^\circ$ angle made by arc $BC$ similarly $\angle BMA = \angle BCC = 60^\circ$ therefore $\angle BMC = \angle BMA$ and that shows $BM$ bisects $\angle AMC$

(2) look at the isoscles triangle $AMD$ with one base angle $AMD = 60^\circ$ that makes $AMD$ equilateral triangle.

(3) triangles $ABD$ and $ACM$ are congruent because $AB = AC, \angle ABD = \angle ACM, \angle BAD= \angle CAM = 60^\circ - \angle DAC$ therefore $$BD = MC, BM = BD + DM = AM + MC$$

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$(3)$ is a consequence of Ptolemy's theorem, from which: $$ MB\cdot AC = MA\cdot BC + MC\cdot AB. $$ Since $AC=BC=AB$, $$ MB = MA+MC $$ follows.

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Hint: Try to prove that $\Delta ABD \cong \Delta ACM$ using parts $(1),(2)$ of the problem.

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