Today, I’m going to write up a quick geometric proof. Here’s the original puzzle that inspired it.

Given that \(AC = AB\) and that \(AE = DE = CD = BC\), what is the measurement of \(\angle A\)?

In other words, the diagram has four isosceles triangles: \(\Delta AED\), \(\Delta EDC\), \(\Delta CBD\), and \(\Delta ABC\).

My strategy for solving the given puzzle was to assign letters to the angles and use three properties to solve for the target variable. These properties are:

- Triangle Sum Theorem: The sum of the angles in any triangle is 180°.
- Linear Angles Theorem: The sum of any set of angles that form a straight line (i.e., linear angles) is 180°.
- The base angles of an isosceles triangle (that is, the two angles opposite the congruent sides) are congruent.

First, I’ll label the diagram, keeping the third property in mind. From here, we can write a set of algebraic equations:

- \(2a + b = 180°\)
- \(2c + d = 180°\)
- \(2e + f = 180°\)
- \(a + c + f + e = 180°\)
- \(b + c = 180°\)
- \(a + d + e = 180°\)

The first four statements come from the Triangle Sum Theorem, and the other two statements come from the Linear Angles Theorem.

Our goal is to solve for \(a\).

Since the large triangle is isosceles, we can conclude that \(c + f = e\) and that therefore \(2e + a = 180°\).

Based on (1) and (5), we can see that: \[2a + b = b + c \\ 2a = c\] Hence, using (2), we get \(2(2a) + d = 4a + d = 180°\). We combine this with (6) to get: \[a + d + e = 4a + d \\ 3a = e\] and, combining with \(2e + a = 180°\), we get: \[2(3a) + a = 6a + a = 7a = 180°\] and so \(a = \frac{180}{7}°\).

For the specific case, then, \(m\angle A = \frac{180}{7}° \approx 25.71°\).

### General Case

Playing around with this, I discovered a pattern. If you draw an isosceles triangle such that it can be broken up into smaller isosceles triangles by drawing \(m\) lines, then \(m\angle A = \frac{180}{3 + 2m}°\). I calculated this for the first four cases (actually, I also did the case where \(m = 0\), but that’s trivial enough to be done mentally: \(m\angle A = 60°\) because it’s equilateral).

I wanted to prove it. To do so, I used a different strategy. Instead of taking a full diagram and setting up equations, as I did above and with the other cases, I started with the assumption that \(m\angle A\) has to be some value, and then built up nesting triangles from there.

Specifically, assume that \(m\angle A = \frac{180°}{k}\). Let’s again call this angle \(a\) and write this as \(ak = 180°\).

Take the first triangle to be formed. Again, we’re focussing on the smaller triangles inside the large circumscribed triangle. The angles of this first triangle are \(a, a, \) and \(180° – 2a = ak – 2a = a(k – 2)\).

Take the next triangle to be formed. We can calculate the base angles because one forms a linear pair with the non-base angle of the triangle we just worked with. That is, the base angles are each \(180° – a(k – 2) = ak – a(k – 2) = 2a\). The non-base angle is \(180° – 2(2a) = ak – 4a = a(k – 4)\).

Take the next triangle to be formed. Now the isosceles angles form a linear triplet (a, d, and e above, for instance), where the base angle (e in the diagram above) = \(180° – a – a(k – 4) = ak – a – ak + 4a = 3a\). The third angle of this triangle is \(180° – 2(3a) = ak – 6a = a(k – 6)\).

At this point, we can prove that, as each triangle is added, the bases angles will measure \(na\) and the third angles will measure \(a(k – 2n)\).

However, at some point, we’ll want to stop. The goal is actually to prove that we can calculate \(a\) based on the number of smaller isosceles triangles inside the larger circumscribing one.

Consider the diagram. We can see that the circumscribing triangle has two base angles. One will have the measure \(na\) because it will be the base angle of the \(n\)th triangle. The other will have the measure \(a(k – 2n) + a(n-1) = a(k – n – 1)\) because it consists of the non-base angle of the \(n\)th triangle and the base angle of the preceding triangle. If the circumscribing triangle is isosceles, then these two angles will have the same measure, that is, \(na = a(k – n – 1)\) \(\rightarrow n = k – n – 1 \rightarrow k = 2n + 1\).

If you have \(n\) inscribed triangles, then \(m\angle A = \frac{180}{2n + 1}°\). In our original case, \(n = 3\), so \(m\angle A = \frac{180}{7}°\).

When will \(m\angle A\) be an integer? When \(2n + 1\) is a factor of \(180\). The odd factors of \(180\) are \(1, 3, 5, 9, 15,\) and \(45\). \(1\) results in a negative value for \(n\), so the values of \(n\) which are positive and which result in an integer value (in degrees) for the measure of \(\angle A\) are \((1, 2, 4, 7, 22)\), resulting in measurements of \((60°, 36°, 20°, 12°, 4°)\).