TL;DR

Using only elementary geometry (not even trigonometry), prove that angle C in the figure equals the sum of angles A and B.

Using only elementary geometry (not even trigonometry), prove that angle C in the figure equals the sum of angles A and B.

There are apparently dozens of proofs of this so I suggest just showing your solution and others can come up with different solutions. I selected High School because it probably is at least that level for today’s readers. Have fun!

The red triangle replicates the 3×1 triangle in the OP; the blue and green ones replicate the 2×1 triangle. These three triangles contribute one side each to the triangle ##PQR##.

The angle ##\angle PRQ## is a right angle, since it is constructed from the green right angled triangle's ##90-B## angle and the blue triangle's ##B## angle. The sides ##PR## and ##QR## are manifestly equal lengths, so the triangle is right angled isoceles and hence angles ##\angle QPR## and ##\angle PQR## are 45°. From the angle ##\angle QPR##, we can see that they are also equal to ##A+B##, so ##A+B=45°##. ##C## is also 45°, obvious from symmetry. Thus ##C=A+B##.

A Chinese colleague observed that a Chinese child would have spent a lot of time learning to do similar puzzles and would therefore have solved it faster (using geometry) than I did (using vectors). But was the time spent learning to solve those time well spent? Did it give them maths skills, or just teach them tricks for solving geometry problems that will never come up in any practical situation?

In each case, the triangle constructed in green is similar to the triangle highlighted in brown which has angle B and is added to angle A. The sum is equal to angle C.

1. Add 3 squares and consider the triangle containing angle B:

2. Add 2 (reflected) copies of the triangle:

3. Note, below, that triangle PQR (red) is right-angled (at P) and isosceles, so has angles ##90^{\circ} – 45^{\circ} – 45^{\circ}##.

Then the distance QM is $$\sqrt{2}-\frac{1}{2\sqrt{2}}= \frac{3}{2\sqrt{2}}$$
then the triangle formed from points PQM is similar to that formed with angle A. So angle PMQ is angle A but angle PMQ=A=C-B so A+B=C.

Hermann's formula

Hutton's formula

It would be hard to use this method in the case of Machin's formula because of ## \arctan (1/239) ##.

First, construct a circle of radius ##n+1## centered on the origin and draw the line segment ##y=\large\frac{x}{n}## to intersect it.

Then construct a circle of radius ##n-1## centered on the intersection point.

Then construct a perpendicular line segment from that intersection point to the green circle.

Finally, draw the line segment from the origin to the point where that perpendicular line segment intersects the green circle. It will go through the point ##(1,1)## showing the sum of the angles is ##\large\frac{\pi}{4}##. I believe this works for any integer ##n##.

In the original problem angle A is defined by the ratio ##\large\frac{1}{3}## so ##n=3## so angle B is defined by the ratio ##\large\frac{3-1}{3+1}=\large\frac{1}{2}##.