Proof of the Angle Sum Property of Triangles.
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1 Proof of the Angle Sum Property of Triangles. The following gives a proof the angle sum property of triangles, along with the proofs of the theorems used intermediately. Some of these have been taken back to axioms, postulates and definitions and a few have been left as suggested exercises. 1. To prove: Sum of the angles of a triangle is 2 right angles. Side BC is produced to D. (Euclid s 2 nd Postulate) (1.1) Given a line and a point not on it, there exists a line parallel to the given line through the given point. (1.2) Given (1.2) There exists a straight line CE parallel to AB. (1.3) AC is a straight line that falls on CE and AB. (1.4) A straight line falling on parallel straight lines makes the alternate angles equal to one another (1.5) Given 1.3, 1.4 and 1.5, <ACE = < BAC (1.6) BD is a straight line that falls on CE and AB. (1.7) A straight line falling on parallel straight lines makes a pair of corresponding angles equal. (1.8) Given 1.3, 1.7 and 1.8 <ECD = <ABC (1.9) Given 1.6 and 1.9, The sum of the angles in the triangle = <ABC + <ACB + < <BCA = <ECD + < ACE + <BCA (1.10) Given 1.1, <ECD + < ACE+ <BCA is two right angles. (1.11) Given 1.10 and 1.11, the sum of the angles of a triangle is 2 right angles. (1.12)
2 Note that this proof applies to ANY triangle, not just the sample of triangles that we happen to have measured. This is the essence of a mathematical proof. In this Statements in bold 1.2, 1.5, 1.8 and 1.11 need to be proved. Starting with (1.5).To Prove: A straight line falling on parallel straight lines makes the alternate angles equal to one another. Let AB and CD be a pair of parallel lines, and F is a straight line that falls on them. (2.1) <AGH and <GHD are a pair of alternate angles. (2.2) Contrary to what we wish to prove, Let <AGH not equal to <GHD. (2.3) Let <GHD be the greater of the two. (2.4) Add <CHG to both. (2.5) Given 2.4 and 2.5 <GHD + < CHG is greater than <AGH + <CHG (2.6) If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles. (2.7) <GHD + < CHG is 2 right angles. (2.8) Given 2.6 and 2.8, <AGH + <CHG is less than 2 right angles (2.9) If a straight line falling on two straight lines makes the interior angles on the same side less than two right angles, the two straight lines, if produced indefinitely, meet on that side on which are the angles less than the two right angles. (2.10) Given 2.9 and 2.10 Lines AB and CD meet on the side of A and C. (2.11) This contradicts 1. Therefore assumption 3 is not true. (2.12) It follows from 2.11 that <AGH IS equal to <GHD Hence Proved. Here 2.7 needs to be proved is the fifth postulate.
3 3 (1.8) To Prove: A straight line falling on parallel straight lines makes the corresponding angles equal to one another. Let AB and CD be a pair of parallel lines, and F is a straight line that falls on them. (3.1) <EGB and <GHD are a pair of corresponding angles. (3.2) When a pair of lines intersect, a pair of vertically opposite angles are congruent. (3.3) Given 3.1 and 3.3, <AGH = <EGB (3.4) A straight line falling on parallel straight lines makes the alternate angles equal to one another.(3.5) Given 3.5, <AGH = <GHD (3.6) Given 3.4 and 3.6, <EGB = <GHD (3.7) Hence proved. Statement 3.5 now is an already proved theorem, which can be used. Statement 3.3 needs to be proved. 4(1.2) To Prove: Given a line and a point not on it, there exists a line parallel to the given line through the given point. BC is a line and A is a point that is not on it. (4.1) Take a point D on BC and join AD. (4.2) Construct <DAE such that it is equal to <ADC. (4.3)
4 {That this is possible can be proved Euclid proposition 23} Extend EA to EF. {2 nd Postulate} (4.4) If a straight lines falls on a pair of lines such that a pair of alternate interior angles are congruent, then the pair of lines are parallel. (4.5) Given 4.2, 4.3, 4.4 and 4.5, EF is parallel to BC. Hence proved. Statement 4.5, which is the converse of Theorem 2 proved above needs to be proved. We proved 1.2, 1.5 and 1.8 of proof 1 above. In the process we see that we have used a few more unproved results. We will prove 2 of these When a pair of lines intersect, a pair of vertically opposite angles are congruent. If a straight lines falls on a pair of lines such that a pair of alternate interior angles are congruent, then the pair of lines are parallel. 5(3.3)To Prove: When a pair of lines intersect, a pair of vertically opposite angles are congruent. If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles. (5.1) AE is a straight line that stands on CD. (5.2) Given (5.1) and (5.2) <AEC + <AED = 2 right angles. (5.3) Similarly, CE stands on AB. (5.4) Given (5.1) and (5.4) <AEC + <CEB = 2 right angles (5.5) Given (5.3) and (5.5)
5 <AED = <CEB (5.6) Similarly it can be shown that <AEC = <DEB (5.7) Hence Proved. 6.(4.5) To Prove: If a straight lines falls on a pair of lines such that a pair of alternate interior angles are congruent, then the pair of lines are parallel. AB and CD are a pair of lines and EF falls on them such that <BEF = <CFE. (6.1) We need to prove that AB is parallel to CD Contrary to what we need to prove, assume, Let AB not parallel to CD. (6.2) If (6.2) is true, AB and CD meet when extended. Let them meet on the side of B and D at G say. (6.3 ) E,F,G form the vertices of a triangle. (6.3) Given 6.1, <AEF an exterior angle of a triangle CFE = <FEG an interior opposite angle. (6.4) But, In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles. (6.5) (6.4) contradicts (6.5) Therefore assumption 6.2 cannot be true. Therefore AD is parallel to CD. Hence proved. A few questions 1) In proof 6, we have yet another unproved result namely - In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles. Is the following an acceptable proof for this result? Why or why not? Proof: The sum of the angles of a triangle is two right angles. (1)
6 Given 1, <ABC + <BCA + <CAB = 2 right angles. (2) If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles. (3) Given 2, <BCA + <ACD = 2 right angles. (4) Given 2 and 4 <ABC + <CAB = <ACD (5) <ABC and <CAB are both positive quantities given ABC is a triangle (6) Given 5 and 6, <ACD is greater than both <ABC and <CAB. Hence proved. 2) Further exploration: You could try to prove the following statements which have been used in the proof. If a straight line stands on a straight line, then it makes either two right angles or angles whose sum equals two right angles. It is possible to construct an angle equal to a given angle on a given straight line and a point on it. In any triangle, if one of the sides is produced, then the exterior angle is greater than either of the interior and opposite angles. In the process, you may use a few more theorems which require to be proved or you may use only axioms, definitions and postulates that are accepted as true.
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