I intended to split my study time today between analytic geometry review and logic review, but ended up getting drawn into some exploration as well. The result is that I have two proofs about quadrilaterals ready to post. The first of the two appears below. It’s a proof of my hunch from the post Squares Again that if a convex, four-sided, equilateral polygon has one right angle, then it has four right angles. It turns out that I could prove this without knowing that such a figure must be a parallelogram. (It must, though. That’s the proof for tomorrow.)
Lemma: The angles opposite (i.e. subtended by) the equal sides of an isosceles triangle are also equal.
Consider the isosceles triangle $\triangle ABC$ with $AB=AC$.
Draw a line bisecting $\angle CAB$ and extend it to intersect $\overline{BC}$. Let the point of intersection be called $D$.
Now, by the side-angle-side property, $\triangle ACD\cong\triangle ABD$, since $AB=AC$, $\angle CAD\cong\angle BAD$, and $\overline{AD}$ is shared.
Therefore, $\angle ACD\cong\angle ABD$. These are the angles opposite the equal sides of the original isosceles triangle, so the lemma is proven.
Proposition: If a convex, four-sided, equilateral polygon has one (interior) right angle, then it has four (interior) right angles.
Let $ABCD$ be an convex, four-sided, equilateral polygon, and let $\angle BCD$ be a right angle.
Divide the polygon by drawing $\overline{BD}$.
Note that, since $\overline{BD}$ is shared and all other line segments are equal, $\triangle ABD\cong\triangle BCD$ by the side-side-side property.
Hence, since $\angle BCD$ is a right angle and the angles of congruent triangles are equal, $\angle BAD$ is also a right angle.
Similarly, $\angle ADB\cong\angle CBD$.
Furthermore, by the lemma, since $\triangle ABD$ is isosceles, $\angle ADB\cong\angle ABD$, and since $\triangle ACD$ is isosceles, $\angle BDC\cong\angle CBD$.
Therefore, $\angle ADB\cong\angle ABD\cong\angle BDC\cong\angle CBD$.
Let the measure of those angles be called $x$. Since the sum of the measures of the angles of a triangle is 180 degrees, either triangle gives us the relationship $90+2x=180$, so $x=45$.
The measures of $\angle ADC$ and $\angle ABC$ are both $2x=2(45)=90$. Thus they are also right angles.
Hence, if a convex, four-sided, equilateral polygon has one right angle, then it has four right angles.
This proof uses SSS again—I actually thought of it while trying to work out a proof of SSS—as well as SAS, through the lemma1. It also uses the fact that the measures of the angles of a triangle sum to 180 degrees. I’m pretty sure that theorem is only a couple of steps from the parallel postulate, though.
My original version of this proof didn’t make use of the lemma about isosceles triangles, but instead argued that the two congruent triangles could be mapped onto each other by either rotation or reflection, and that was why all of the acute angles had to be equal. That was kind of a nifty argument, but not easy to formalize.
Stay tuned tomorrow for the proof about parallelograms.
- To do. ↩︎
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