## 311 – HW 8

April 23, 2015

HW 8 is due Tuesday, April 28, at the beginning of lecture.

Work in hyperbolic geometry. Given a triangle $\triangle ABC$, recall that its Saccheri quadrilateral $\Box ABB'A'$ based at $\overleftrightarrow{AB}$ is defined as follows: Let $M$ be the midpoint of $\overline{AC}$ and $N$ be the midpoint of $\overline{CB}$. Let $A',B'$ be the feet of the perpendiculars from $A$ and $B$ to $MN,$ respectively.

Continuing with the same notation, suppose now that $G$ is an arbitrary point on $\overleftrightarrow{MN}$, and let $H$ be a point on the ray $\overrightarrow{AG}$ with $GH=AG$. Show that $\Box ABB'A'$ is also the Saccheri quadrilateral of $\triangle ABH$ based at $\overleftrightarrow{AB}$.