Theorem 1: In a circle, if an angle that opens on the diameter, then it is a right angle.
Proof:
(1) Let E be the center of the circle.
(2) BE ≅ BA ≅ CE since they are all radii.
(3) Since triangle AEB and triangle AEC are isoceles triangles (see here if needed), we can conclude (see here) that:
∠ ABE ≅ ∠ BAE
∠ ACE ≅ ∠ CAE
(4) And step #3 gives us that ∠ BAC = ∠ ABC + ∠ ACB.
(5) But we also know that ∠ FAC = ∠ ABC + ∠ ACB since:
(a) ∠ FAC = 180 degrees - BAC [Angles of a straight line add up to 180 degrees, see here if needed]
(b) ∠ ABC + ∠ ACB = 180 degrees - BAC [Angles in a triangle add up to 180 degrees, see here if needed]
(6) And since ∠ FAC ≅ ∠ BAC, both must be right angles [since 2*x = 180 degrees → x = 90 degrees]
QED
Postulate 1: Similar segments of circles on equal straight lines equal one another.
Euclid originally presented this postulate as a theorem using the principle of superposition (see here for details). I am presenting it as a postulate in order to avoid the superposition.
Lemma 1: In equal circles, angles stand on equal circumferences whether they stand at the centers or the circumferences.
Proof:
(1) Let ABC and DEF be congruent circles with ∠ G ≅ ∠ H and ∠ A ≅ ∠ D.
(2) Since they are congruent, all radii are congruent so that:
BG ≅ CG ≅ EH ≅ FH
(3) So we have triangle BGC ≅ triangle EHF by side-angle-side (see here if needed)
(4) From step #3, we know that BC ≅ EF
(5) So that segment BAC ≅ segment EDF. [See Postulate I above]
(6) And this implies that segment BKC ≅ segment ELF.
QED
Lemma 2: In a circle, the angle at the center is double the angle at the circumference when the angles have the same circumference as base.
Proof:
(1) Let ABC be a circle with center E.
(2) EA ≅ EB since both are radii.
(3) ∠ EAB ≅ ∠ EBA since the base angles of an isoceles triangle are congruent (see here for details if needed).
(4) ∠ BEF = ∠ EAB + ∠ EBA (since angles of a triangle add up to 180 degrees and since two angles of a straight line add up to 180 degrees) so ∠ BEF is double ∠ EAB.
(5) We can use the same line of reasoning to establish that ∠ FEC is double ∠ EAC.
(6) Putting step #4 and step #5 together gives us that ∠ BEC is double ∠ BAC
(7) We can use this same reasoning to prove that ∠ GEC is double ∠ EDC.
(8) We can also prove that ∠ GEB is double ∠ EDB.
(9) Therefore the remaining ∠ BEC is double ∠ BDC.
QED
Theorem 2: In equal circles, angles standing on equal circumferences are equal to one another, whether they stand at the center or at the circumference.
Proof:
(1) So, we can assume that circumference BC ≅ circumference EC
(2) Assume ∠ BGC ≠ ∠ EHF
(3) Then one of them is greater; let's assume ∠ BGC is greater (if ∠ EHF is greater, we can make the same argument in terms of EHF)
(4) Construct ∠ BGK equal to ∠ EHF on the straight BG and at the point G on it (see here for details on the construction).
(5) Now equal angles stand on equal circumferences when they are at the centers, therefore circumference BK equals circumference EF (see Lemma 1 above)
(6) But EF equals BC so therefore BK equals BC which is a contradiction since BK is smaller than BC.
(7) So, we reject our assumption and conclude that ∠ BGC ≅ ∠ EHF
(8) The angle at A is half of the angle BGC. [See Lemma 2 above]
(9) The angle at D is half of the angle EHF [See Lemma 2 above]
(10) Therefore, the angle at A also equals the angle at D.
QED
Lemma 3: A line tangent to a point on a circle forms a right angle with a line drawn from that point to the center of the circle.
Proof:
(1) Assume that ∠ FCD is not a right angle
(2) Let FG be a line that is perpendicular to DE
(3) So triangle FGC is a right angle with the hypotenuse at FC.
(4) So FC is greater than FG by the Pythaogrean Theorem [See the Corollary here for details]
(5) But FC = FB since radii are congruent.
(6) So FC is less than FG which contradicts step #4.
(7) So we have a contradiction and we reject our assumption.
QED
References:
- David Joyce, Euclid's Elements, Book III, Proposition 31
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