1. The Elements, Book I – Propositions 11 – 16 MONT 104Q – Mathematical Journeys: Known to Unknown September 28, 2015

2. “Erecting” a perpendicular Proposition 11. To construct a line at right angles to a given line from a point on the line. Construction is closely related to Proposition 10: Given point A on the line, use Postulate 3 to construct two other points on the line B, C with AB = AC. Construct an equilateral triangle ΔBCD (Proposition 1) Then DA is perpendicular to the line at A.

3. “Dropping” a perpendicular Proposition 12. To drop a perpendicular to a given line from a point not on the line. Construction: Given point A not on the line and P on the other side of the line, use Postulate 3 to construct a circle with radius AP and center A – it intersects the line in points B, C with AB = AC. Let D be the midpoint of BC (Proposition 10) Then DA is perpendicular to the line at D. Proof: ΔADB and ΔADC are congruent by Proposition 8 (“SSS”). Hence <ADB = <ADC are right angles. QEF

4. A group of propositions about angles Proposition 13. If from a point on a line a ray is drawn, then this ray forms with the line two angles whose sum is the same as two right angles. Proof: Say the ray starts at point B on the line, P,Q are on the line on opposite sides of B and A is on the ray. If <PBA = <QBA then the two angles are right angles (given as a Definition by Euclid). Otherwise, use Proposition 11 to erect a perpendicular to the line at B, and take C on the perpendicular. Then reasoning with Common Notions 1 and 2, <PBA + <QBA = <PBC + < QBC so equal to two right angles. QED

5. A group of propositions about angles, continued Proposition 14. If two angles have a side in common, and if the noncommon sides are on different sides of the common side, and if the angles are together equal to two right angles, then the noncommon sides lie along the same straight line. This is a converse of Proposition 13. The reasoning is similar in that it is based just on the Common Notions. Note: Euclid did not consider 180˚ (“straight”) angles as angles – the equivalent for him was the angle described by two right angles together – not a huge difference, of course, but it affected the way a number of statements were stated and proved.

6. A group of propositions about angles, continued Proposition 15. Vertical angles are equal. Note: these are the opposite angles formed by the intersections of two lines – like <CPD and <APB:

7. A group of propositions about angles, continued Proof: <BPC + <CPD is the same as two right angles by Proposition 13. Similarly for <APB + <BPC. Hence <CPD + <BPC = <APB + < BPC by Postulate 4. Therefore, <CPD = <APB by Common Notion 3. QED

8. A group of propositions about angles, continued Proposition 16. In a triangle, an exterior angle is greater than either of the nonadjacent interior angles. The statement is that <DCB is greater than <CAB, <CBA:

9. Proof of Proposition 16 Euclid's proof is clever! To show <DBA is greater than <BAC: Construct the midpoint E of AC (Proposition 10) and extend BE to BF with BE = EF (Postulate 2 and Proposition 3). Construct CF (Postulate 1). Note that <AEB = <FEC by Proposition 15.

10. Proof of Proposition 16, concluded Therefore ΔAEB and ΔCEF are congruent (Proposition 4 – “SAS”). Hence <BAE = <ECF. But <ECF is a part of the exterior angle <DCA. So the exterior angle is larger (Common Notion 5). QED A similar argument shows <DCA is larger than < ABC.