Tag: 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. ↩︎

Today I worked on exercises in Analysis with an Introduction to Proof and did a little mathematical puttering.

I’m also ready today to start sharing my work on the proposition about the inscribed equilateral triangle. I’ll start by proving that a triangle is equilateral if and only if its three angles are equal.

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First, recall from my post Revenge of the Squares, Part 1 that the angles opposite (i.e. subtended by) the equal sides of an isosceles triangle are also equal. The proof of the following related lemma is very similar.

Lemma: If two angles of a triangle are equal, then the sides that subtend them are also equal.

Consider the triangle $\triangle ABC$ with $\angle ACB\cong\angle ABC$.

Draw a line bisecting $\angle BAC$ and extend it to intersect $\overline{BC}$. Let the point of intersection be called $D$.

Now, by the angle-angle-side property, $\triangle ACD\cong\triangle ABD$, since $\angle ACB\cong\angle ABC$, $\angle CAD\cong\angle BAD$, and $\overline{AD}$ is shared.

Therefore, $AC=AB$. $\overline {AC}$ and $\overline {AB}$ are the sides that subtend the equal angles $\angle ACB$ and $\angle ABC$, so the lemma is proven.

(Note that I have added AAS to my list of triangle congruence properties to prove.¹)

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Proposition: A triangle is equilateral if and only if its three angles are equal.

Consider a triangle $\triangle ABC$.

Assume that $\triangle ABC$ is equilateral.

Then, since $AB=AC$, $\angle ACB\cong\angle ABC$, by the theorem cited above. Similarly, since $AC=BC$, $\angle ABC\cong\angle BAC$.

It follows that $\angle ACB\cong\angle ABC\cong\angle BAC$.

Thus, if $\triangle ABC$ is equilateral, then its three angles are equal.

Assume instead that the three angles of $\triangle ABC$ are equal.

Then, by the lemma, since $\angle ACB\cong\angle ABC$, $AB=AC$. Similarly, since $\angle ABC\cong\angle BAC$, $AC=BC$.

It follows that $AB=AC=BC$.

Thus, if the three angles of $\triangle ABC$ are equal, then it is equilateral.

This proves the equivalence.

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Stay tuned tomorrow for some circle geometry.

1. To do. ↩︎