on gspon gspGiven ray AC and ray AB going in opposite directions such that the two rays lie on the line BC and share endpoint A, show angle CAB equals 180º. From the definition of betweeness, since A belongs to the line segment BC, and A is not equal to B or A is not equal to C, then A is between B and C . From this fact we know that the three points B, A, and C are on the same line. So we know that the three points A, B, and C are distinct but collinear points. Now consider the degenerate triangle CAB. By the SAS Congruence Axiom, we know that triangle CAB is negatively congruent to itself using side CA and side BA, which are congruent to themselves by reflexivity, and angle A, which the two triangles share . So, a(CAB) =- a(CAB). And therefore, 2*a(CAB) = 0º = 360º. Thus, a(CAB) = 180º. And so, given a straight line, we know that the angle is equal to 180º. menu
Proof:
1) Here we are given that angle CAB is equal to 180º in
both the triangle above and the degenerate triangle below. We
wish to show that the triangle is congruent to the degenerate
triangle in order to prove that the 180º angle is in fact
on a straight line.
2) Because these two triangles are the same, we can say that by the SAS Congruence Theorem that segments CA and AB are congruent to themselves by reflexivity, and angle A is shared by both triangles.
3) Therefore, the above triangle is equal to the degenerate triangle below, and because it was given that angle CAB is equal to 180º, the same angle CAB on the degenerate triangle is also equal to 180º. So, if an angle is equal to 180º, then it is in fact a straight angle.
Given a triangle with two congruent sides, the angles opposite those congruent sides are also congruent.
Given a triangle with two congruent sides, prove the triangle has two congruent angles. Note to Dr. McCrory, we chose not to do a proof similar to the one that we used for the converse of this theorem because we would need to use the SSS congruency axiom and the group that proved SSS existed used this isosceles triangle theorem.
on
gspDefine an isosceles triangle as one which has two congruent sides. Given isosceles triangle ABC with congruent sides AB and BC, show that angle BAC and angle BCA are congruent. Define angle BAC as opposite to side BC, angle BCA as opposite to side AB, and angle ABC as opposite side AC. By the angle completeness axiom, a(ABC) = x mod 360 for some real number x. Since x is an element of the reals, x/2 is also an element of the reals. Construct a ray BE, between ray BA and ray BC such that a(ABE)= x/2 mod 360. By the angle addition axiom a(ABE) + a(CBE) = a(ABC) = x mod 360. Then x/2 mod 360 + a(CBE) = x mod 360, thus a(CBE) = x/2 mod 360. Hence, a(ABE)= x/2 mod 360 = a(CBE), giving us congruent angles. Define this process as the bisecting of an angle. Call BE the angle bisector of a(ABC). In other words we may divide an angle into two congruent angles. Since ray BE was constructed between ray BA and ray BC, the ray BE will intersect the side opposite angle ABC. Based on our definition of opposite sides and angles we know ray BE will intersect AC. Choose the point of intersection as the point named E on BE. Thus we have created the triangles ABE and CBE. Since sides AB and BC were given as congruent, angle ABE and angle CBE are congruent, and side BE is congruent to itself by reflexivity. Now, by the side-angle-side congruence axiom, triangle ABE is congruent to triangle CBE. Therefore, because corresponding parts of congruent triangles are congruent, we know angle BAC is congruent to angle BCA. menu
Given two congruent angles contained in a triangle, show that the sides opposite those congruent angles are congruent.
on gspGiven a triangle JKM with a(JKM) congruent to a(JMK) show that JK is congruent to JM. We know KM is congruent to MK by symmetry. So now using the group 4 angle-side-angle congruence thereom, we know that triangle KMJ is congrent to triangle MKJ. Then, because corresponding parts of congruent triangles are congruent, JK is congruent to JM. menu
Given a line AB and a point C not on line AB show that there exists a line perpendicular to AB through point C. Here we define perpendicular to mean two lines which intersect and form one 90º angle. Choose any radius so that a circle centered at point C with that radius intersects line AB at two points, A and E. Then, at points A and E construct congruent circles with radii greater than 1/2 the distance between points A and E. Find the point where the two circles intersect on the opposite side of line AB from point C. Call this point G. Now make segments AC, CE, EG, GA, and CG. We know that AC is congruent to CE because they are radii of congruent circles. Also, EG is congruent to AG because they are radii of congruent circles.
Now consider triangle ACE. From the definition of isosceles triangles, triangle ACE must be isosceles because it was constructed with two congruent sides. Therefore a(CAE) is congruent to a(CEA) since the two angles opposite the congruent sides are also congruent (proven previously). Also triangle AGE is isosceles since it also satisfies the definition. So, we know that a(GAE) is congruent to a(GEA) by the same reasoning as above. By the angle addition axiom, a(CAG) = a(CAE) + a(GAE) and a(CEG) = a(CEA) + a(GEA). Since a(CAE) is congruent to a(CEA) and a(GAE) is congruent to a(GEA), then substituting gives us a(CAG) = a(CAE) + a(GAE) = a(CEA) + a(GEA) = a(CEG). We can now say that triangle CAG is congruent to triangle CEG by the SAS axiom of congruence. Therefore, since a(ACG) and a(ECG) are corresponding parts of congruent triangles we know that they are congruent. Calling the point where line AB and segment CG intersect H we can look at triangle ACH and triangle ECH. From the line intersection rule of construction we know that the two segments will have a point of intersection, H. Because CH is a shared side of the two triangles, a(ACG) and a(ECG) are congruent, and AC is congruent to CE we know that triangle ACH and triangle ECH are congruent by SAS axiom of congruence. Therefore since a(AHC) and a(EHC) are corresponding angles in congruent triangles, the angles must be equal. Also, a(AHC) and a(EHC) make up a straight line which is comprised of 180º by our proof of the number of degrees in a straight angle. Therefore the two angles have the same measurement and a sum of 180º, so we know that they each must be 90º. Thus line AB and segment CG are perpendicular. menu
Given line AB and point C on line AB. Show that there exists a perpendicular line to line AB through point C. Here we define perpendicular to mean two lines that intersect and form one 90º angle. Choose a radius q so that a circle from point C with radius q intersects AB at two points. We know these points exists from the circle-line intersection rule. Call these two points H and I. Now using H and I as centers, with any radius r greater than the radius q. These two circles intersect at two points which we know exist from the circle intersection rule. Choose one of the two points and call it point J. Now construct segments HJ, IJ, HC, and CI. We know HC is congruent to IC because they are radii of the same circle. Also, we know HJ is congruent to IJ because they are radii of congruent circles. Now consider triangle HIJ. It is an isosceles triangle based on the definition of isosceles triangles. Therefore, a(IHJ) is congruent to a(HIJ) since the two angles opposite congruent sides are congruent ( as previously proven). So now by the side-angle-side congruence axiom, triangle CHJ and triangle CIJ are congruent. Then because corresponding parts of congruent triangles are congruent a(HCJ) and a(ICJ) are congruent. Also, a(HCJ) and a(ICJ) make up a straight line which is comprised of 180º by our proof of the number of degrees in a straight angle. Therefore the two angles have the same measurement and a sum of 180º, so we know that each must be 90º. Thus line AB and segment CJ are perpendicular. menu
Given a segment AB, show that the perpendiuclar line to AB such that for any point X on the perpendicular line d(AX) = d(BX) is the perpendicular bisector of AB. From the distance axiom, we have that line segment AB, between A and B is the set of all points C such that d(A,C) + d(C,B) = d(A,B). If C belongs to the line segment between A and B, but C is not equal to A or B, we say that C is between A and B. Choose C so that d(AC) = 1/2 d(AB). Now by substitution we have 1/2 d(AB) + d(CB) = d(AB), so d(CB) = 1/2 d(AB). So d(AC) = 1/2 d(AB) = d(CB). Call this point C the midpoint. Now, from the perpendicular existence theorem, we know that there exists a perpendiucular line to AB through point C on segment AB. Choose any point X on the line perpendicular to AB. Construct segments AX, BX, and CX. Consider triangle AXC and triangle BXC. We know that a(XCB) = 90º = a(XCA) from the definition of a perpendicular. We also know that XC is congruent to itself by reflexivity and that d(AC) = d(CB). Therefore by the side-angle-side congruence axiom triangle AXC is congruent to triangle XBC. Because corresponding parts of congruent triangles are congruent, AX is congruent to BX. So, d(AX) = d(BX). Since the choice of X was random, it must hold true for all points on the line XC, which is the perpendicular bisector of AB. menu
Given a segment AD and a line through AD such that for all points C on that line, d(AC) = d(DC). We will show that this line is the perpendicular bisector of AD. Consider the triangle ACD. From the definition of betweeness, we know that there exists a point E that lies on segment AD such that E is not equal to A or D. By the angle addition axiom, a(ACD) = a(ACE) + a(DCE). Also, by the angle completeness axiom, a(ACD) = x mod 360 for any real number x. Since x is an element of the reals, x/2 is also an element of the reals. Therefore, given a segment AD, there exists some point E between A and D such that E does not equal A or D, and d(AE) + d(ED) = d(AD), from the definition of betweeness. So, we create point E between A and D such that a(ACE)= x/2 mod 360. By the angle addition axiom a(ACE) + a(DCE) = a(ACD) = x mod 360. Then x/2 mod 360 + a(DCE) = x mod 360, thus a(DCE) = x/2 mod 360. Hence, a(ACE)= x/2 mod 360 = a(DCE), giving us congruent angles. Now consider triangle ACE and triangle DCE. By the side-angle-side congruence axiom, triangle ACE is congruent to triangle DCE since CE is congruent to itself by reflexivity. Therefore, because corresponding parts of congruent triangles are congruent, d(AE) = d(DE) and a(CEA) = a(CED). Also, a(CEA) and a(CED) make up a straight line which is comprised of 180º by our proof of the number of degrees in a straight angle. Therefore the two angles have the same measurement and a sum of 180º, so we know that each must be 90º. Thus, CE is the perpendicular bisector of AB. menu