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.
Very fascinated that you posted this triangle. I have been working very closely with it in relation to tetrahedra. Also, impressed with your proof. Angle ABD is twice that of ACD.
Imagine a book that has equilateral triangles as its front and back covers and only one page that is an equilateral triangle as well. Opening the book so the covers are flat on the table, the page defines a two complementary tetrahedra as it is lifted. A side view of this “book” is the triangle that you have brought up. If we have a diameter of 2(sqrt 3), the radius is sqrt 3. The edge of the triangle is then 2. As the page turns, 5 of either tetrahedron edges remain fixed, and only one edge varies per tetrahedron.
Not sure if you care to hear about Buckminster Fuller’s Synergetic Geometry, but he argued the use of the regular tetrahedron to be the unit of measure. So, at the point where the page is defining the sixth length and it is also 2, just as the other 5, we have the regular tetrahedron. The complementary tetrahedron is 1/4 of a regular octahedron (variable edge = 2(sqrt 2). Fuller states that the regular tetrahedron with its edges being equal to 2, has the volume of 1 tetrahedral units, 3/4(sqrt 2) cubic units. This gives most sheer anger in such a statement and there is the conversion factor.
Most interesting, is if the page is at the vertical point the two variable edges are the same (sqrt 6). The cubic volume is 1 cubic unit. Yes, that tetrahedron has a volume of 1. The conversion factor between the two is sqrt 9/8 to go from cubic to tetrahedral units, and sqrt 8/9 the other way.
There is a series of volumes that go from the vertical point of the page all the way down to a cover. The volumes can be stated as having a denominator of either 9 for cubic and 8 for tetra units. The numerator is just the integers descending from 9 on down. So, sqrt 9/9 = 1 cubic unit and 8/8 = 1 tetrahedral unit.
When the numerator is 4, the lenghts AD = 2(phi^-1) or 1.236068 and CD = 2(phi^1) or 3.236068. This is of great interest to me, which I will not get into.
Lastly, if one had a tesselation of equilateral triangle and they all had this book cover and page scenario, an incidental or inadvertent tetrahedron is produced to fill all-space. It would have 4 edges of 2, one of one varying length and the other the complementary varied length.
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