# In triangle $ABC$, $\angle A = 20º$ and $AM = CN = CB$. Find angle $MBN$.

Recently, I stumbled across this question which had some interesting geometry questions from a primary school competition. Here is one of the questions:

In the diagram below, $$ABC$$ is an isosceles triangle, where $$\angle A = 20^{\circ}$$. Let $$M$$ and $$N$$ be two points on $$AC$$ such that $$AM = CN = CB$$. What is the measure, in degrees, of $$\angle MBN$$?

After drawing this diagram on paper, I realised that simple angle chasing would not do. $$CN = CB$$ means that $$\Delta CBN$$ is isosceles of course, but I was having trouble connecting this to side $$AM$$. I also did not want to use trigonometry and yet I could not see a way to apply theorems such as similar triangles, circle theorems or power of a point. I then used GeoGebra and gradually worked my way up to this diagram (link):

I noticed a few curious things when I constructed this diagram. It turned out that $$BC = BD$$ as well, and in turn $$BD = MD$$ where $$D$$ is constructed such that $$AM = MD$$ and $$D$$ lies on $$AB$$. Then I realised in a flash of lighting that I could make use of $$AM = CN = CB$$ and reflect the triangle where the base $$AB$$ is constant. Yet more odd things popped up: the circle passing through $$B', C'$$ and centred at $$A$$ also passed through $$M$$; $$ND$$ was parallel to $$CB$$.

Using all of these insights, I am very close to a solution but not there yet.

We draw an equilateral triangle with side $$AB$$ as shown. Then $$\triangle PAC$$ is isosceles with $$\angle PAC = 40^\circ$$. That leads to $$\angle BPC = 10^\circ$$. Also, $$\angle PBC = 20^\circ$$

Now $$\triangle PBC \cong \triangle BAM$$ (by S-A-S)

So it follows that $$\angle ABM = 10^\circ$$.

$$\therefore \angle MBN = 80^\circ - 50^\circ - 10^\circ = 20^\circ$$

Using angle chasing, $$\angle ACB = \angle ABC = \frac{180º-20º}{2} = 80º$$. Since $$\Delta NCB$$ is isosceles, $$\angle BNC = \angle NBC = 50º$$.

Now we have to make use of the information $$AM = CN = CB$$, focusing in on $$AM$$ in particular. Thus, let us reflect triangle $$ABC$$ horizontally around the middle, so that base $$AB$$ stays in the same position. Thus $$CN = CB = AC' = AM$$. Since $$\angle B' = 80º$$ by symmetry, $$\angle CAM = 80º - 20º = 60º$$. Since $$\Delta CAM$$ is also iscoceles, this triangle must be equilateral! Therefore $$\angle AC'M = \angle AMC' = 60º$$ as well.

Again by symmetry, we have that $$AE = EB$$. Thus $$\angle EBA = 20º$$ also and $$\angle AEB = 140º$$. Now if we construct $$D$$ such that $$AM = MD$$ and $$D$$ lies on $$AB$$, then $$\angle MDA = 20º$$ as well. Thus by AA similarity, $$\Delta AMD \sim \Delta AEB$$.

• I think meant $\triangle C'AM$ being equilateral? Mar 13, 2022 at 12:12
• Yep, thanks for reading over my work and spotting the typo. Mar 13, 2022 at 13:36
• ∠C′AM=80º−20º=60º Mar 13, 2022 at 23:48