use Venema 0.6.3 that in order to explain why CD is the perpendicular bisector of AB in the picture capture explain what has to be prove 0.6.3. The perpendicular bisector of a segment AB is a line € such that l intersects ABat its midpoint and l IAB. Prove the following theorem.Pointwise Characterization of Perpendicular Bisector. A point P lies on the per-pendicular bisector of AB if and only if PA = PB.OP OQProof : (forward direction) Suppose point P lies onthe perpendicular bisector ofOur goal is to show PA = PB.Notice the triangles AAMP and ABMPCongruent "by SAS withincluded angles at M. Thus thecorresponding sides PA and PB arecongruent and PA = PB.segment AB.areAMB(backward direction) Given segment AB andpoint P Such that PA= PB, we want toshow P is on the perpendicular bisector ofAB. Let M be the midpoint of AB.Consider line PM. We Know it is a bisectorof AB and want to show the anglesZ AMP and BMP are right. Thetriangles AAMP and ABMP are congruentmidpoint, PM is common,by Šss (since M wasand PA = PB by hypothesis). Thus thecorresponding angles Question 2 ProofGiven a line segment, constructing a midpoint with Euclid propositions.We will show that we can find the midpoint of a line segment using straightedgeand compass. In Euclid geometry, we know that a Midpoint is a point in the exact middleof a given straight line segment. We will be proving this using Proposition 1 that is Toconstruct an equilateral triangle on a given finite straight line. We will also be referring toProposition 9 to bisect a given rectilinear angle. Putting these propositions together we willbe able to construct a midpoint to any given segment.CE = 6.58IEAE = 3.8BE = 3.8ED = 6.58Given Segment AB. We can draw two circles, one with center A and radius AB,the other with center B and radius BA. When drawing the two circles we created two pointsof intersection that is where the circles cross points C and D. Then if we draw a straight lineconnecting the two created intersection points they cross through the line segment AB. Thepoint of intersection is called E which we will show is the midpoint.

we have to prove CD is perpendicular bisector of AB we will use Venema 0.6.3, According to which A…

Juestion 2 Proot Given a line segment, constructing a midpoint with Euclid propositions. We will show that we can find the midpoint of a line segment using straightedge and compass. In Euclid geometry, we know that a Midpoint is a point in the exact middle of a given straight line segment. We will be proving this using Proposition 1 that is To construct an equilateral triangle on a given finite straight line. We will also be referring to Proposition 9 to bisect a given rectilinear angle. Putting these propositions together we will be able to construct a midpoint to any given segment.

Juestion 2 Proot Given a line segment, constructing a midpoint with Euclid propositions. We will show that we can find the midpoint of a line segment using straightedge and compass. In Euclid geometry, we know that a Midpoint is a point in the exact middle of a given straight line segment. We will be proving this using Proposition 1 that is To construct an equilateral triangle on a given finite straight line. We will also be referring to Proposition 9 to bisect a given rectilinear angle. Putting these propositions together we will be able to construct a midpoint to any given segment.

Trigonometry (MindTap Course List)

8th Edition

ISBN:9781305652224

Author:Charles P. McKeague, Mark D. Turner

Publisher:Charles P. McKeague, Mark D. Turner

Chapter1: The Six Trigonometric Functions

Section: Chapter Questions

Problem 2GP

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