Section 2 - Euclid's Axiomatic Approach

Question 1

what is Euclid’s parallel axiom?

Answer 1

If a straight line crossing two straight lines makes the interior angles on one side together less than two right angles, then the two straight lines will meet on that side.

Question 2

what is Playfair’s axiom?

Answer 2

given a line l and pt P (not on l), there is exactly one line on P // to l.

(this is equivalent to Euclid’s parallel axiom)

Question 3

what does 𝙸.32 state?

Answer 3

the sum of the angles in a △ is 2 right angles

Question 4

how do we prove the sum of the angles in a △ is 2 right angles?

Answer 4

• let ABC be a △
• draw a line DE through C // to AB
• ∠BAC=∠DCA & ∠ABC=ECB
• since ∠DCA+∠ECB+∠ACB = 180 degrees, ∠ACB+∠CBA+∠BAC also = 180 degrees

Question 5

what is the SAS axiom?

Answer 5

if △s ABC & A’B’C’ have |AB|=|A’B’|, ∠ABC=∠A’B’C’ and, |BC|=|B’C’|, then |AC|=|A’C’|, ∠BCA=∠B’C’A’ and, ∠CAB=∠C’A’B’

we can use this to prove SSS and ASA

Question 6

what does 𝙸.5 state?

Answer 6

In isosceles triangles the angles at the base equal one another, and, if the equal straight lines are produced further, then the angles under the base equal one another

Question 7

how do we prove the base angles in an isosceles△ are equal?

Answer 7

• suppose △ABC has |AB|=|BC|
• △ABC & △CBA are congruent by SAS
• the base angles are equal

Question 8

what is a parallelogram?

Answer 8

quadrilateral in which opposite sides are //

Question 9

how do we prove that opposite sides of a //-gram are equal?

Answer 9

• let ABCD be a //-gram
• draw AC
• ∠DAC=∠ACB & ∠CAB=∠DCA
• by ASA △ACD=△CAB
• then |AD|=|CB| & |CD|=|AB|

Question 10

what does 𝙸.15 state?

Answer 10

vertically opposite ∠s are equal

Question 11

what does 𝙸𝙸.4 state?

Answer 11

If a straight line is cut at random, then the sum of the rectangles contained by the whole and each of the segments equals the square on the whole.

e. g. (a+b)² = a² + 2ab + b²
e. g. a(b+c) = ab +ac

Question 12

AAA gives …

Answer 12

△s are congruent

Question 13

AAS gives …

Answer 13

△s are same

Question 14

ASA gives …

Answer 14

△s are same

Question 15

SAS gives …

Answer 15

△s are same

Question 16

SSA gives …

Answer 16

△s are same only if A is right angle

Question 17

SSS gives …

Answer 17

△s are same

Question 18

area of a //-gram is same as ….

Answer 18

area of rectangle with same base

Question 19

how do you convert a rectangle into a //-gram with same base & height?

Answer 19

• given □OPQR & ST on extension of PQ
• |ST|=|OR|=|PQ|
• area of //-gram OSTR= area OPTR - area △OPS
• by SSS △OPS=△RQT
• so area □OPQR = area □OPQR + area △RQT - area △OPS
• so area □OPQR = area //-gram OSTR

this argument is reversible

Question 20

//-grams with same …….. have same area

Answer 20

base & height

Question 21

since area of //-gram is bh, what it area of△?

Answer 21

1/2 bh

Question 22

what is the written version of pythagoras?

Answer 22

in any right angle △ the sum of the squares on the legs equals the square on the hypotenuse

Question 23

what is Thales’ thm? how can we restate it?

Answer 23

//s cut any lines they cross into proportional segments
or
a line drawn // to one side of a △ cuts the other two sides proportionally

Question 24

how do we prove that a line drawn // to one side of a △ cuts the other two sides proportionally?

Answer 24

• given △ABC
• let P be on AB & Q be on AC s.t. PQ//BC
• draw BQ & PC
• then area △PQC = area △QPB (same base & height)
• so area △ABQ = area △ACP
• now area △ABQ = area △APQ + area △PBQ
• △APQ & △PBQ have same height
• so |AP|/|PB| = area△APQ/area△PBQ
• similarly |AQ|/|QC| = area△APQ/area△PQC = area△APQ/area△QPB
• then |AP|/|PB| = |AQ|/|QC|

Question 25

how do we prove that if A&B are two pets on a circle then for all pts C on one of the arcs joining them, ∠ABC is constant?

Answer 25

• let 0 be the centre of the circle, with pts A,C,B
• draw OA, OB, OC, AC, & BC
• know |OA|=|OB|=|OC| (radii)
• then ∠OAC=∠OCA & ∠OCB=∠OBC (isosceles △s)
• so ∠AOC = 180 - 2∠OCA & ∠BOC = 180 - 2∠OCB
• then ∠AOB= 360 - 2(∠OCA+∠OCB)
• but ∠AOB is a constant independent of C so ∠OCA+∠OCB=∠ACB is a constant

Question 26

how do we find √x for any constructible length x

Answer 26

use the corollary that if AB is a diameter of a circle, and C is a pt on the circle, then √ACB is a right angle

• take a pt D on AB s.t. |AD|=x & |DB|=1
• then bisect AB at D and extend this line to the circle
• call this pt C
• then |DC|=√x by similar △s

Question 27

how can we find the centre of a circle?

Answer 27

• take any 4 pts A,B,C,D on the circle
• draw lines AB & CD and bisect them
• the perpendicular bisectors will intersect at the centre of the circle