I ran into a problem at a meeting of teachers, yesterday, which exposed an embarrassing hole in my geometrical knowledge — and so I quickly became obsessed with filling it. In the diagram, the large triangle is right, and the leg lengths were given; the problem was to find the length of the hypotenuse (also the diameter of the circle centered at B). The median seen here was not shown, however, and no right angle was identified. Were the triangle not a right triangle, this would be an impossible problem, so I knew it had to be a right triangle . . . but that didn’t satisfy me. I had to have a proof, so I wrote one.

Here it is: in the diagram shown, segment AC is a diameter of a circle with center B, while D is any point on the triangle distinct from A and C. Segments BA, BD, and BC are all radii of the same circle, and therefore have the same length, making triangles ABD and CBD isosceles with bases, respectively, of AD and CD.

Let the measure of angle ABD be some number x. Since it forms a linear pair with angle CBD, angle CBD’s measure must be 180 – x.

Angles BAD and BDA are the base angles of isosceles triangle ABD, which has a vertex angle measure already chosen as x. Since these base angles must be congruent, it follows from the triangle sum theorem that each of these angles must measure (180 – x)/2.

Angles BCD and BDC are the base angles of isosceles triangle CBD, which has a vertex angle measure already determines to be 180 – x. Since these base angles must be congruent, it follows from the triangle sum theorem that each of these angles must measure (180 – (180 – x))/2.

By the angle sum theorem, the measure of angle ADC must equal the sum of the measures of angles BDC and BDA, already shown, respectively, to be (180 – (180 – x))/2 and (180 – x)/2.

Angle ADC’s measure therefore equals (180 – (180 – x))/2 + (180 – x)/2, which simplifies to (180 – 180 + x)/2 + (180 – x)/2, which further simplifies to x/2 + (180 – x)/2. Adding these two fractions yields the sum (x + 180 – x)/2, and then the “x”s cancel, leaving only 180/2, or 90 degrees, for the measure of angle ADC. Therefore. triangle ADC, the large triangle in the diagram, must be a right triangle — QED.

I’m rather embarrassed that I didn’t already know this property of inscribed triangles with one side being the diameter of the triangle’s circumscribed circle — but at least I figured the proof out myself, and that, in turn, made the faculty meeting easily the least boring one I have ever attended.