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I'm trying to go through the section of Newer Results in Hartshorne's Geometry: Euclid and Beyond.

This particular exercise has been bugging me for a good while: enter image description here

My first approach was to show that the perpendicular bisectors of $A'B'$ and $A'D'$ intersect at the same point as the perpendicular bisectors of $A'B'$ and $B'C'$. I figured since the center of the circle circumscribed around a triangle has its center at the intersection of the three perpendicular bisectors, this would show that $A'$, $B'$, $C'$ and $D'$ would all be on the same circle. However, I didn't see any way to show this.

I also tried extending $A'D'$ and $A'C'$ down into the circle containing $D$, $D'$ and $C$ with hopes that it $\angle D'A'C'\cong\angle D'B'C'$, but this seemed like a dead end also.

I'm going a little mad, if any one sees a possible solution, I'd be quite grateful. Thanks.

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Dear yunone, you might want to clarify (maybe in the title) that the "Hartshorne" book you are referring to is not the (in)famous "Algebraic geometry" (which is what initially sprang to mind for me at least). – Akhil Mathew Feb 20 '11 at 1:50
@Akhil, done! I'm sure his Algebraic geometry is quite beyond me. – yunone Feb 20 '11 at 1:54
up vote 2 down vote accepted

I see two ways of doing it. The first is probably more in spirit of classical geometry, the second uses the cross ratio.

  1. Let $\beta = \angle A'B'C'$ and $\delta = \angle A'D'C'$. We want to show that $\beta + \delta = \pi$. Draw the sides of the five quadrilaterals $AA'B'B$, $BB'C'C$, $CC'D'D$ and $DD'A'A$ as well as $ABCD$ into the picture. By assumption, these quadrilaterals are cyclic, hence opposite angles are supplementary (add up to $\pi$). We have \begin{align*} \beta & = 2\pi - \angle A'B'B - \angle BB'C' = (\pi - \angle A'B'B) + (\pi - \angle BB'C') \\ & = \angle A'AB + \angle C'CB. \end{align*} Similarly, $\delta = \angle A'AD + \angle C'CD$. Therefore \[ \beta+\delta = (\angle A'AB + \angle A'AD) + (\angle C'CB + \angle C'CD) = \angle DAB + \angle DCB = \pi \] as we wanted.

  2. Compute the cross ratio $[A',B',C',D']$ and show that it is real by using that the cross ratios $[A,B,C,D]$ and $[A,A',B',B]$, $[B,B',C',C]$, $[C,C',D',D]$ and $[D,D',A',A]$ are all real.

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Thank you Theo, I completely forgot that opposite angles of cyclic quadrilaterals are supplementary. A very nice proof indeed! – yunone Feb 20 '11 at 3:16
@yunone: It was my pleasure! When I had to do these things in high school, I never really got the hang of it. However, it seems like I should finally have a look at this "Baby-Hartshorne" (no offense!), it looks really nice. Keep posting these problems, I like them. – t.b. Feb 20 '11 at 3:35

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