Plane Geometry - Bruce E. Shapiro

SECTION 47. PERPENDICULAR LINES IN HYPERBOLICGEOMETRY 273Figure 47.3: The alternate interior angles δ = γ iff P Q is a common perpendicularto l and m and P M = QM.Theorem 47.3 (Uniqueness of Common Perpendiculars) If two linesadmit a common perpendicular that perpendicular is unique.Proof. (Exercise.)Theorem 47.4 Suppose that lines l and m are parallel and are cut by acommon transversal. Then the alternate interior angles are congruent ifand only iff the following two conditions hold: (1) lines l and m admit acommon perpendicular; (2) the transversal intersects the common perpendicularsegment at its midpoint.Proof. (⇒)Suppose δ = γ (see figure 47.3). If δ = 90 and γ = 90 thenthe result of the theorem follows immediately. Hence we assume that theseangles are not right angles.Let M be the midpoint of RS, and drop perpendiculars from M to m and l,calling their feet Q and P as shown in fig 47.3. By AAS △SQM ∼ = △RP M.Hence ∠SMQ = ∠P MR. Thus −−→ MP and −−→ MQ are opposite rays thereforeP M and P Q form a single segment P Q.By congruence of triangles, P M = QM, and P M is a common perpendicular.(⇐) Assume that the transversal t intersects a common perpendicular QPat the its midpoint M, as shown in figure 47.3. We need to show that∠δ = ∠γ.Since ∠QMS = ∠P MR by the vertical angle theorem, triangles △MQS ∼ =△MP R by ASA.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.