Question

I need help doing a doing two column for these two propositions.

Book 1 Proposition 7:

Given two straight lines constructed from the ends of a straight line and meeting in a point, there cannot be constructed from the ends of the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively, namely each equal to that from the same end.

Book 3 Proposition 14:

Equal straight lines in a circle are equally distant from the center, and those which are equally distant from the center equal one another.

This is one of my classmates example on hers.

Untitled_Edited Helvetica Regular BI 1.0 tE v 1) In any triangle, the side opposite the greater angle is greater 2) In a triangle, the longest side of all sides equates to the the greatest angle diagonal from given side 3) W.T.S: In AABC, segment BC corresponding to AA > segment AC corresponding to AB. 1. In ΔABC, let segment BC > segment AC Given Straight Edge & Compas By construction (2) and det.。 By def. of opposite sidesequals opposite angles By L.T.T and (3) (4) By def. of an exterior angle in a triangle 2. Construct segment AD making D a point on segment BC 3. Segment AC Segment CD 5. AACD is an isosceles triangle 6. In AADB, A ADC is the exterior angle By Exterior Angle Theorem Angle Addition Postulate By Substitution (7) (8) By (9) 10. ACAB >ABC 11. Therefore, Segment BC corresponded to ACAB > Segment AC corresponded to A ABC 4) If we also prove by Contradiction (), for that segment's BC or AA is the greater angle. Let's assume that XABC> AACB, which is given. By Proposition 5 of Euclid's Elements or by the Isosceles Triangle Theorem, the base angles following the anger opposite these sides are so, 114ABC ACB then this cannot be proven true since we know KABC> XACB. Therefore, Case 1(Segment's AC AB) has to be False. In the next case, we have to show Segment's AC<AB. So, we still know that XABC > ^ACB and we can infer that segments AC and AB are two sides of a triangle. By Proposition 18 in Euclid's Elements, or one the Triangle Inequalities(Ihm. 13), states that in any triangle the angle opposite the greater side is greater;- AACB > ^ABC. We also show that this case 2 is proven to be false since it's given KABC> AACB. Finally, we show our last case 3 in that Segment's AC > AB. We assume again that AABC> ^ACB. So, we have eliminated AC < AB, AC £ AB, and we now know Theorem 13 (The angle opposite the longest side is the greatest) BC>AC, we will prove that the angle corresponding to Segment that AC> AB by contradiction. (~-) -Isabella Briguglio (Proposition 19) Math 329

Answer 1

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