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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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    @Akhil, done! I'm sure his Algebraic geometry is quite beyond me.2011-02-20

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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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    @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.2011-02-20