I need to solve the following problem without using trigonometry.

Given $\triangle ABC$ with $\angle C=120^\circ$. Point M is on the side $AB$, such that $\angle CMB=60^\circ$ and $BM:AB=1:3$. Find $\angle B$.

Using law of sines it's easy, but without it seems impossible.

  • 1
    $\begingroup$ You could use $ABC$ instead of $A_1A_2A_3$. $\endgroup$ – ghosts_in_the_code May 19 '15 at 15:52
  • $\begingroup$ I think the labels can be chosen arbitrary $\endgroup$ – parkhyeyoo May 19 '15 at 21:22
  • $\begingroup$ I made the edits, it looks a lot cleaner now. $\endgroup$ – Sawarnik May 20 '15 at 5:40
  • $\begingroup$ Can anyone help with this please? $\endgroup$ – parkhyeyoo May 20 '15 at 16:23
  • $\begingroup$ maybe there is a way to prove it using the circumcircle? $\endgroup$ – parkhyeyoo May 22 '15 at 22:24

Solution to the above problem

Bisect $AM$ at $N$. Construct equilateral $\triangle$s $NME, MNO$. Then $AN=NM=ME=EN=NO=OM=BM$.

Thus $\triangle$s $EOM, OEN$ have a common base $EO$ and all four short sides are equal, so they are congruent (SSS); their obtuse angles are $120^\circ$, so their acute angles are $30^\circ$.

Each of the angles $OMB, BME, ENA, ANO$ is an exterior angle of an equilateral $\triangle$ and is thus $120^\circ$, so the $\triangle$s $OBM, BEM, EAN, AON$ are congruent to the above $\triangle$s $EOM, OEN$ (SSA). Thus $AE=AO=EO=BE=BO$ so $\triangle$s $AOE, BEO$ are equilateral.

Thus $O$ is the centre of the circle through $A, E, B$. $\angle AEB=120^\circ=\angle ACB$, so $C$ lies on this same circle.

$BMN$ is a straight line and $\angle BMC=60^\circ$, so $\angle CMN=120^\circ$; $\triangle MNO$ is equilateral, so $\angle NMO=60^\circ$, so $OMC$ is also a straight line.

$\angle NOA=30^\circ$, so $\angle COA=90^\circ$. Thus $\angle CBA=45^\circ$ as the angle at the circumference is half the angle at the centre.


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