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In the proof of Lemma 1 we mentioned that many in- correct methods for finding a vertex $p$ such that the line segment $b p$ is an interior diagonal of $P$ have bee...

Question

In the proof of Lemma 1 we mentioned that many in- correct methods for finding a vertex $p$ such that the line segment $b p$ is an interior diagonal of $P$ have been published. This exercise presents some of the incorrect ways $p$ has been chosen in these proofs. Show, by considering one of the polygons drawn here, that for each of these choices of $p,$ the line segment bp is not necessarily an interior diagonal of $P$ . a) $p$ is the vertex of $P$ such that the angle $\angle a b p$ is small-

In the proof of Lemma 1 we mentioned that many in- correct methods for finding a vertex $p$ such that the line segment $b p$ is an interior diagonal of $P$ have been published. This exercise presents some of the incorrect ways $p$ has been chosen in these proofs. Show, by considering one of the polygons drawn here, that for each of these choices of $p,$ the line segment bp is not necessarily an interior diagonal of $P$ . a) $p$ is the vertex of $P$ such that the angle $\angle a b p$ is small- est. b) $p$ is the vertex of $P$ with the least $x$ -coordinate (other than $b ) .$ c) $p$ is the vertex of $P$ that is closest to $b$ .



Answers

Assign coordinates to each vertex and write a coordinate proof. Given: $\angle R$ is a right angle in $\triangle P Q R . A$ is the midpoint of $\overline{P R}$. $B$ is the midpoint of $\overline{Q R}$. Prove: $A B=\frac{1}{2} P Q$.


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