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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.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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 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 AB at 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 OQ Proof : (forward direction) Suppose point P lies on the perpendicular bisector of Our goal is to show PA = PB. Notice the triangles AAMP and ABMP Congruent "by SAS with included angles at M. Thus the corresponding sides PA and PB are congruent and PA = PB. segment AB. are A M B (backward direction) Given segment AB and point P Such that PA= PB, we want to show P is on the perpendicular bisector of AB. Let M be the midpoint of AB. Consider line PM. We Know it is a bisector of AB and want to show the angles Z AMP and BMP are right. The triangles AAMP and ABMP are congruent midpoint, PM is common, by Šss (since M was and PA = PB by hypothesis). Thus the corresponding angles <AMP and <BMP are congruent. But they also make a linear pair, So can conclude both LAMP and <BMP we are right angles. So line PM is the perpendicular bisector of AB (and paint P is on Q.E.D. Note: If P=M, then aso P is on the I bisector.
Transcribed Image Text:0.6.3. The perpendicular bisector of a segment AB is a line € such that l intersects AB at 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 OQ Proof : (forward direction) Suppose point P lies on the perpendicular bisector of Our goal is to show PA = PB. Notice the triangles AAMP and ABMP Congruent "by SAS with included angles at M. Thus the corresponding sides PA and PB are congruent and PA = PB. segment AB. are A M B (backward direction) Given segment AB and point P Such that PA= PB, we want to show P is on the perpendicular bisector of AB. Let M be the midpoint of AB. Consider line PM. We Know it is a bisector of AB and want to show the angles Z AMP and BMP are right. The triangles AAMP and ABMP are congruent midpoint, PM is common, by Šss (since M was and PA = PB by hypothesis). Thus the corresponding angles <AMP and <BMP are congruent. But they also make a linear pair, So can conclude both LAMP and <BMP we are right angles. So line PM is the perpendicular bisector of AB (and paint P is on Q.E.D. Note: If P=M, then aso P is on the I bisector.
Question 2 Proof 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. CE = 6.58 IE AE = 3.8 BE = 3.8 ED = 6.58 Given 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 points of intersection that is where the circles cross points C and D. Then if we draw a straight line connecting the two created intersection points they cross through the line segment AB. The point of intersection is called E which we will show is the midpoint.
Transcribed Image Text:Question 2 Proof 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. CE = 6.58 IE AE = 3.8 BE = 3.8 ED = 6.58 Given 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 points of intersection that is where the circles cross points C and D. Then if we draw a straight line connecting the two created intersection points they cross through the line segment AB. The point of intersection is called E which we will show is the midpoint.
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