Unformatted text preview: through P. Proof: Let l be any line and let P be a point not on line l . By the "Drop a Perpendicular" Theorem, there is a unique line t such that t passes through P and l ⊥ t . By the "Drop a Perpendicular" Theorem, there is unique line m such that m passes through P and t ⊥ m . Therefore, l and m share the common perpendicular t . Q E D P l t P l m t P l...
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- Spring '10
- Quantification, Euclidean geometry, Axiom