Show that triangle $ABC$ is isosceles.

the question

Consider the triangle $$ABC$$ and a point $$M$$ inside the triangle such that $$\angle MAB = 10 ,\angle MAC = 40 ,\angle MCA = 30$$ and $$\angle MBA = 20$$ . Show that triangle $$ABC$$ is isosceles.

my idea

the drawing

As you can see I calculated the other angles we have in the triangle...I'm preatty sure we need an auxiliar construction, but I can't figure out which one to do. I dont know how to start. Hope one of you can help me! Thank you!

• Extending $BM$ you get an exterior angle of triangle $BMC$ equal to $80^\circ$. This means $\angle MBC + \angle MCB = 80^\circ$. This and the boomerang theorem on $MACB$ gives $\angle AMB = \angle MAC + \angle ACB + \angle MBC$. Try using this to your advantage. Apr 18 at 13:41
• Later today, I will return here to leave a complete proof of this problem without trigonometry. Apr 18 at 13:57
• @Rusurano sadly properties of triangles was left out of our syllabus, hence I didn't get the chance to look into it in depth. My answer was nothing but trigo. I am looking forward to your answer :)
– Gwen
Apr 18 at 14:32
• Are you familiar with the trigonometric form of ceva's theorem? Apr 18 at 16:28
• @Gwen Posted, enjoy Euclidean geometry! :) Apr 19 at 6:47

By law of sines, $$\frac{AM}{\sin 30°}=\frac{AC}{\sin 110°} \tag{1}\label{1}$$ $$\frac{AB}{\sin 150°}=\frac{AM}{\sin 20°} \tag{2}\label{2}$$ We can combine ($$\ref{1}$$) and ($$\ref{2}$$) to get $$\frac{AC\sin 30°}{\sin 110°}=\frac{AB\sin 20°}{\sin 150°}$$ Now we can apply values of $$\sin 30°$$, $$\sin 150°$$: \begin{align} AC& =2AB \left(2\sin 20°\sin 110°\right) \\ AC & =2AB(\cos(-90°)-\cos(130°)) \\ AC & =-2AB\cos(180°-50°) \\ AC & =2AB\cos 50° \tag{3} \label{3} \end{align} Again by law of cosines, $$BC^2=AC^2+AB^2-2ABAC\cos 50°$$ Now substituting from $$(\ref{3})$$, \begin{align} BC^2& =4(AB)^2\cos^2 50° + AB^2 - 4(AB)^2\cos^2 50° \\ BC^2 & =AB^2 \\ BC & =AB \end{align} Hence $$\triangle ABC$$ is isoceles.

• How did you know that $2*sin(20)*sin(110)=cos(-90)-cos(130)$ Apr 18 at 14:23
• That follows an identity $$\cos(A-B)-\cos(A+B)=2\sin A\sin B$$ It can be easily proven by expanding them
– Gwen
Apr 18 at 14:25

Here is a more general hint towards solving these kinds of problems. When you see a problem where degrees are given in $$10x$$ increments (this is sometimes disguised a bit), the problem is likely constructed with the following schematic in mind, where an equilateral triangle is divided by rays subtending $$10$$ degrees each. Note the emergence of many right angles, and several $$80-20-80$$ isosceles triangles. These latter triangles are special, for, wherever two congruent $$80-20-80$$ triangles share an $$80$$-degree angle, embedded within them are several equilateral and isosceles triangles with vertices on the intersection of the sides of one and altitude of the other triangle. It's a good exercise to prove this yourself. Note that this only works for congruent, not similar, pairs of such triangles (see the blue vs gray equilateral triangles I marked).

In your particular version, you are given a situation with side $$AB$$ and point $$M$$ (in red) and the ray $$BC'$$ (also in red). You only do not know where the final side is, but it is instantly fixed once you find the triangle at $$M$$ with an angle of $$150$$, for this is part of the $$80-20-80$$ in the lower right corner of the diagram with $$MF$$ as its upper side.

• This is actually beautiful approach and definitely deserves an upvote. Apr 19 at 11:14
• just a typo: "and the ray $AC′$ (also in red)" - ray $BC'$ is in red. Apr 25 at 16:16
• @PiotrWasilewicz - thanks, fixed Apr 25 at 17:00

This solution was given by user vineet on AoPS.

Construct an equilateral triangle $$\triangle AMD$$ with base $$AM$$, on the opposite side of $$B$$. Let $$E = DB\cap CM$$.

Claim 1: $$BM$$ is the perpendicular bisector of $$AD$$.
Proof: $$\angle AMB = 150^\circ$$, so $$BM$$ is angle bisector of $$\angle DMA$$, hence the perpendicular bisector of $$AD$$.

Claim 2: $$BD \perp AC$$
Proof: Due to claim 1, $$A$$ and $$D$$ are symmetric about $$BM$$. Particularly, $$\angle BDM = \angle BAM = 10^\circ$$. Also $$\angle CAD = 20^\circ$$ (as $$\angle DAM = 60^\circ$$) and $$\angle ADM = 60^\circ$$. Summing them, we get the required claim.

Claim: $$AMED$$ is cyclic.
Proof: Due to claim 2, $$\angle DEC=60^\circ = \angle DAM$$.

Then, note that, $$\angle EAM = \angle EDM = \angle BDM = 10^\circ$$, so that $$\angle EAC = 30^\circ = \angle ECA \implies \triangle ECA \text{ is isosceles}$$This, coupled with claim 2, gives that $$BD$$ is the perpendicular bisector of $$AC$$, which proves the required statement. $$\blacksquare$$

Here is an elementary proof of this statement that does not involve trigonometry.

By plane separation axiom, line $$AC$$ separates a plane into two half-planes. Construct an equilateral triangle $$CAK$$ on side $$AC$$, so that point $$K$$ lies in the same half-plane with point $$B$$.

Extend a segment $$CM$$ to intersection with $$AK$$ at a point $$H$$. As $$CH$$ is the angle bisector of $$\angle ACK$$, and $$_\Delta ACK$$ is equilateral by construction, then $$CH$$ is also a median and an altitude.

Drop a perpendicular $$KL$$ on $$AC$$. As $$_\Delta ACK$$ is equilateral, it will also be the median and the angle bisector, so $$\angle AKL = \angle CKL = \frac{1}{2}\cdot 60^\circ = 30^\circ$$.

In $$_\Delta AMB$$, $$\angle AMB = 180^\circ - (\angle MAB + \angle ABM) = 180^\circ - (10^\circ + 20^\circ) = 180^\circ - 30^\circ = 150^\circ$$.

In right triangle $$_\Delta AMH$$, $$\angle AMH = 90^\circ - \angle MAH = 90^\circ - 20^\circ = 70^\circ$$. Therefore, $$\angle BMH = \angle AMB - \angle AMH = 150^\circ - 70^\circ = 80^\circ$$. $$\angle OMB$$ is supplementary to $$\angle BMH$$, therefore $$\angle OMB = 180^\circ - \angle BMH = 180^\circ - 80^\circ = 100^\circ$$.

In right triangle $$_\Delta KOH$$, $$\angle KOH = 90^\circ - \angle OKH = 90^\circ - 30^\circ = 60^\circ$$.

Warning: the following part contains a circular argument. If you know how to fix it, please do!

Assume that point $$B$$ does not belong to $$KL$$. Then let $$B_1$$ be the intersection point of $$MB$$ and $$KL$$.
Consider the triangle $$_\Delta AB_1K$$. There, $$\angle B_1AK = 10^\circ$$, $$\angle AKB_1 = 30^\circ$$. Therefore, $$\angle AB_1K = 180^\circ - (\angle B_1AK + \angle AKB_1) = 180^\circ - 40^\circ = 140^\circ$$.
Now, consider the triangle $$_\Delta MOB_1$$. There, $$\angle B_1OM = 60^\circ$$, $$\angle B_1MO = 100^\circ$$. By exterior angle theorem, an exterior angle should be equal to the sum of two opposite interior angles. $$\angle MB_1K$$ is exterior to $$_\Delta MOB_1$$. Therefore, $$\angle MB_1K = \angle B_1OM + \angle B_1MO = 60^\circ + 100^\circ = 160^\circ$$.
Since $$AB_1K = 140^\circ$$ and $$\angle MB_1K = 160^\circ$$, we also must have $$\angle AB_1M = \angle MB_1K - \angle AB_1K = 160^\circ - 140^\circ = 20^\circ$$.
$$\angle ABM = 20^\circ$$. Then we have $$\angle AB_1M = \angle ABM$$. But by angle construction axiom, there is only one way to construct an angle less than $$180^\circ$$ on a given half-plane. So we have a contradiction. Therefore, points $$B$$ and $$B_1$$ are coincident.

As point $$B$$ belongs to $$KL$$, and $$KL$$ is a perpendicular bisector of $$AC$$, then, by the properties of perpendicular bisector, point $$B$$ is equidistant from the points $$A$$ and $$C$$. Therefore, $$AB = AC$$, which leads to the conclusion that $$_\Delta ABC$$ is isosceles by definition, which was to be proven.

By the way, this is Problem 5 from USAMO 1996, and there are pretty trigonometric solutions published on Art of Problem Solving.

• This is wrong. $\angle B_1AK$ is not $10^\circ$. $\angle BAK$ is. Here, you have assumed $\angle B_1AK=10^\circ$ and hence your argument is circular.
– D S
Apr 19 at 7:19
• @DS Ah, true. Though, similar reasoning must absolutely work if I define $B_1$ correctly. Would you suggest a fix to that reasoning? Apr 19 at 8:16
• Your proof uses a circular argument and yet it received four votes. Standards are deteriorating. Apr 26 at 20:56

We can construct an isosceles triangle $$AFC$$ by drawing a segment $$AF$$ at a $$10^\circ$$ angle to $$AM$$. $$FF_\perp$$ is the perpendicular bisector of this triangle with $$\angle F_\perp F C = 60^\circ$$. This means that the angle formed between $$F F'$$ (the extension of $$F_\perp F$$) and $$FM$$ is also $$60^\circ$$. Since $$\angle AMC=110^\circ$$ it is also the case that $$\angle AFM=60^\circ$$ and so segment $$FM$$ bisects $$\angle AFF'$$.

If we can prove that the point where the extension of $$F_\perp F$$ and $$AB$$ intersect is vertex $$B$$ then $$BF_\perp$$ is a perpendicular bisector of $$AC$$ and therefore $$\triangle ABC$$ is isosceles with $$|AB|=|BC|$$.

Lets call the point of intersection of lines $$AB$$ and $$F_\perp F$$, $$B_F$$. $$\triangle AFB_F$$ has $$M$$ as a concurrent point for segments $$AM$$, $$FM$$ and $$B_FM$$. We know that $$AM$$ (resp. $$FM$$) bisects $$\angle B_FAF$$ (resp. $$\angle AFB_F$$). The only point of concurrency with this property is the incenter which means that $$B_FM$$ also bisects $$\angle AB_FF$$. Since $$\angle AB_FF=40^\circ$$ it follows that $$\angle AB_FM=20^\circ$$. But we know from the conditions of the problem that $$\angle ABM=20^\circ$$. This is only possible if $$B \equiv B_F$$ and this completes the proof.

HINT.-Let $$a,b,c$$ the sides. The angles to determine are $$\alpha=\angle{MCB}$$ and $$\beta=\angle{MBC}$$. These are solution of the system

$$\alpha+\beta=80^{\circ}\\ac\sin 50^{\circ}=bc\sin (20+\beta)=ab\sin(30+\alpha)$$ there are two solutions $$(\alpha, \beta)=(20,60)$$ which yields to $$b=c$$ and $$(\alpha, \beta)=(50,30)$$ which yields to $$b=a$$

• Can you elaborate on how you solved this system? There seems to be way too many unknowns, like why can't we let $\alpha = 10^\circ, \beta = 70^\circ$, and set $a, b, c$ accordingly? Apr 19 at 15:16