constructing proving a square Proposition 46 from Book

constructing & proving a square Proposition 46 - from Book 1 of The Elements [Bonus material: proving the regular pentagon!] Brought to you by: François, Frank & Zack

Euclid: Book 1, Proposition 46 “To Construct A Square With A Given Side”

Constructing the Square

Step 1 Let AB be the given straight line (Postulate 1). A B

Step 2 Draw AC at right angles to given line AB from point A (Proposition 11). C A B

Step 3 Make a point D on line AC so that AD=AB (Proposition 3). C D A B

Step 4 Draw DE from point D parallel to AB (Proposition 31). C E D A B

Step 5 Draw BE from point B Parallel to AD (Proposition 31). C D E A B

Step 6 and Step 7 Since ADEB is a parallelogram, AB=DE and AD=BE (Proposition 34). Since in “Step 2” we said lines AD=AB, then lines AD=DE=AB=BE make parallelogram (Common Notion 1). C E D A B ADEB an equilateral

Step 8 Since line AD falls upon parallels AB and DE, BAD+ ADE=2 right 29). < < angles (Proposition C E D A B

Step 9 Since BAD is a right angle as stated in “Step 1”, < ADE must also be another right angle. < C E D A B

Step 10 In “parallelogrammic” areas, the opposite sides and angles are =, so each of the opposite angles of ABE and BED are also right. As a result, ADEB is right angled (Proposition 34). < < C E D A B

Step 11 Since we proved ADEB is equilateral and right angled, it is therefore a square, and it is described along line AB (Definition 22). C E D A B

The Pentagon! (Book IV, Proposition 11) Given: Circle ABCDE F A **Where triangle FGH is isosceles and where angles G and H are each double that size of angle F. (Book 4, Prop 10) E B C D G H

(1) Inscribe triangle FGH into the circle, where triangle ACD is equiangular with triangle FGH (Book IV Prop. 2) - F A E B C D G H

Bisect the angles ACD and ADC using straight lines CE and DB. A Angles ACD and CDA are double the size of angle CAD. E B -But they have been bisected, therefore these five angles, DAC, ACE, ECD, CDB and BDA are all equal to each other. C D

Circumferences Equal angles stand on equal circumferences , so all five circumferences are equal (AB, BC, CD, DE, EA). (Book III, Prop. 26) AB, BC, CD, DE, EA are equal A E B C D

Equilateral as Lines AB, BC, CD, DE, and EA are equal. Straight lines that cut off equal circumferences are equal, so lines AB, BC, CD, DE, and EA are equal. (Book III, Prop. 29) Therefore pentagon ABCDE is equilateral. A E B C D

Equiangular Circumference AB = DE Add BCD to each so: AB + BCD = DE + BCD (CN 2) A Therefore ABCD = EDCB. Angle AED is on circumference ABCD, and angle BAE is on circumference EDCB (Book III, Prop 27) therefore angle BAE = angle AED. Same applies for the rest of the angles so all angles are equal and it is Equiangular. E B C D

Equilateral Pentagon + Equiangular Pentagon = Regular Pentagon

THE END