Month: April 2024

More low-level depression and concentration problems today, but I did do a bit of watching and reading. I also finished writing up the proof below, which I know you’ve all be waiting for. As I explained yesterday, I’ve tried to make it a bit more readable than other proofs I’ve posted recently.

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Proposition: If $\triangle ABC$ is an equilateral triangle inscribed in a circle and $D$ is a point on the circle between $A$ and $C$, then $AD+CD=BD$.

Place a point $E$ on $\overline{BD}$ so that the distance from $D$ to $E$ is the same as the distance from $D$ to $C$, as shown below.

Draw the segment $\overline {CE}$.

Since the original triangle, $\triangle ABC$, is equilateral, its three angles must be equal by the proposition from the post Easy Lemmas.

The sum of the measures of the three angles of a triangle is $\pi$ radians. Thus, when the three angles are equal, each one must measure $\frac{\pi}{3}$. It follows that the angles of $\triangle ABC$ have this measure: $\angle BAC=\angle ABC=\angle ACB=\frac{\pi}{3}$.

Because they are both subtended by arc $BC$, by the second proposition from the post Angles and Arcs, $\angle BAC$, the leftmost of the angles of the original equilateral triangle, is equal to $\angle BDC$, the rightmost of the angles with their verticies at $D$.

Similarly, because they are both subtended by arc $AB$, $\angle ACB$, the rightmost of the angles of the original equilateral triangle, is equal to $\angle ADB$, the leftmost of the angles with their verticies at $D$.

It follows that the two angles with their verticies at $D$, $\angle BDC$ and $\angle ADB$, both measure $\frac{\pi}{3}$.

Consider the newly formed triangle $\triangle CDE$. Two of its sides, $\overline{CD}$ and $\overline{DE}$, are equal by construction (i.e. because of the way point $E$ was chosen). Thus, the two angles subtended by those sides, $\angle CED$ and $\angle DCE$, are equal to one another by the lemma from the post Revenge of the Squares, Part 1.

Furthermore, since $\angle CDE$, the other angle of $\triangle CDE$, is one of the angles that measures $\frac{\pi}{3}$, it follows that $\angle CED$ and $\angle DCE$ together must measure $\pi-\frac{\pi}{3}=\frac{2\pi}{3}$ and that each one individually must measure $\frac{\pi}{3}$.

Hence the three angles of $\triangle CDE$ are all equal, from which it follows, by the proposition in the post Easy Lemmas, that $\triangle CDE$ is equilateral.

Now consider $\angle BEC$, the angle supplementary to $\angle CED$, the uppermost of the angles of $\triangle CDE$. Since the angles of $\triangle CDE$ all measure $\frac{\pi}{3}$, the supplementary angle $\angle BEC$, must measure $\frac{2\pi}{3}$.

Note that, since each one measures $\frac{\pi}{3}$, the sum of $\angle ADB$ and $\angle BDC$, the two equal angles with their verticies at $D$, is also $\frac{2\pi}{3}$. Thus, the measure of $\angle ADC$, the large angle they form, is also $\frac{2\pi}{3}$.

Yet that means that the two triangles highlighted above in yellow, $\triangle BCE$ and $\triangle ACD$, are congruent by the side-side-angle property, which applies when the angle in question is greater than or equal to $\frac{\pi}{2}$. The triangles’ large angles, $\angle BEC$ and $\angle ACD$, are equal, as just shown, and measure $\frac{2\pi}{3}$; their long sides, $\overline {BC}$ and $\overline {AC}$, are equal, since both are sides of the original equilateral triangle $\triangle ABC$; and another pair of their sides, $\overline {CE}$ and $\overline {CD}$, are equal because both are sides of the new, smaller equilateral triangle $\triangle CDE$.

It follows that the remaining sides, $\overline {BE}$ and $\overline{AD}$, are equal, as well. Hence $BE=AD$.

As above, $DE=CD$ by construction.

Thus $AD+CD=BE+DE$. Yet $\overline {BE}$ and $\overline {DE}$ together make up the larger segment $\overline {BD}$. Therefore $AD+CD=BD$ and the theorem of is proven.

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With the addition of SSA for right and obtuse angles, I think I might have a complete set of triangle congruence properties to prove.¹

I could almost have done this proof without assigning measures to the angles. Equality between $\angle BEC$ and $\angle ADC$ could be established using the property of triangles illustrated below, and I think I could have shown that they are obtuse as well. Without angle measures, though, I couldn’t think of a way to show that, with two sides of $\triangle CDE$ equal by construction, not only must the angles they subtended be equal to one another, but both must be equal to the third angle.

Let me know what you thought of this verbose-mode proof. I found it a little difficult to write without an audience clearly in mind.

1. To do. ↩︎

My back is somewhat better today. I spent an hour on exercises in Analysis with an Introduction to Proof and I plan to read some more of Infinite Powers this evening, as well.

I’m also ready to prove the proposition about inscribed angles from my post Groundwork. It turns out that proposition falls right out of another that reader Tim McL suggested starting with when I said I knew little about the geometry of circles.

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Proposition: If a central angle and an inscribed angle are subtended by the same arc of a circle, then the measure of the central angle is twice that of the inscribed angle.

Let $A$ and $B$ be points on a circle with center $D$.

Draw the central angle subtended by arc $AB$.

Draw an inscribed angle also subtended by arc $AB$ and let its vertex be called $C$.

Join $\overline {CD}$.

Now suppose that the sides of the two angles intersect only at $A$ and $B$ (as in the figure above).

Since $\overline{BD}$ and $\overline{CD}$ are both radii of the circle, $BD=CD$, from which it follows that $\angle CBD = \angle BCD$ (as I’ve proven previously).

Similarly, $\overline{AD}$ and $\overline{CD}$ are both radii of the circle, so $AD=CD$ and $\angle CAD = \angle ACD$.

Futhermore, since the sum of the angles of a triangle is $\pi$ radians, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.

It is also true that $\angle BCD + \angle ADC + \angle ADB = 2\pi$.¹

Combining these equations will yield $\angle ADB = 2(\angle BCD + \angle ACD)$.

But $\angle ACB = \angle BCD + \angle ACD$, so $\angle ADB = 2\angle ACB$, and the theorem is proven for this case.

Suppose instead that sides of the angles intersect other than at $A$ and $B$, as shown above. (Note that this argument does apply in the edge case where $\overline{BC}$ intersects $\overline{AD}$ at $D$.)

Then, $\angle ACB =\angle ACD-\angle BCD$ and $\angle ADB = \angle BDC – \angle ADC$.

As above, $2\angle BCD + \angle BDC = \pi$ and $2\angle ACD + \angle ADC = \pi$.

Combining these equations will yield $\angle ADB = 2(\angle ACD – \angle BCD)$.

Hence $\angle ADB = 2\angle ACB$, and the theorem is proven for this case, as well.

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After proving this proposition, I looked up how Euclid did it. His solution was largely the same. It was slightly more elegant, using the property of triangles illustrated below to cut out some of the algebra, but not so elegant as to dispense with the cases.

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Proposition: If two inscribed angles are subtended by the same arc of a circle, then they are equal.

I’m just going to sketch a proof of this. By the proposition above, the blue and green inscribed angles subtended by arc $AB$ both have half the measure of the red central angle subtended by the same arc. Therefore, they must be equal to each other.

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Footnotes

1. Do I need to prove this? I have an idea of how I would do it, but it seem so fundamental. ↩︎