Overview of Section 2.1 Basic Properties of Arcs and Central and Inscribed Angles
Congruent circles -- iff equal radii.
Arc of a circle
Measure of the minor arc is the measure of the central angle.
Propositions. Note, these are intuitive statements that could be proved, but since the proofs are omitted they are called 'propositions.'
Proposition 2.1: Two minor arc in the same circle or in congruent circles are congruent if and only if their central angles are congruent.
Proposition 2.2: In a circle, or in congruent circles, two arcs are congruent if and only if the corresponding chords are congruent.
Proposition 2.3: A diameter perpendicular to a chord bisects the chord and bisects each of the arcs determined by the chord
Proposition 2.4: If two circles intersect, the line through their centers is the perpendicular bisector of their common chord.
A proof was given for Proposition 2.4.
Tangent to a Circle
Tangent to a curve: A tangent at a point P on a curve . . . is a line containing P and having the same direction of the curve at P. If Q is another point on the curve, PQ is a secant. As Q gets close to P from either side, PQ approaches the tangent line to the curve at P. Note we could have a secant PQ that is also a tangent line.
Definition of a Tangent to a Circle: A tangent to a circle O at a point P on the circle is the line through P, perpendicular to the radius OP.
NOW SOLVE THIS 2.1
Prove: The tangent to the circle intersects the circle in exactly one point.
OPEN GSP FILE
An Angle Formed by a Tangent and a Secant
This section discusses a tangent and secant having a common point P on the circle.
Theorem 2.1: The measure of the angle formed by a tangent and a chord (secant?) measures 1/2 of the intercept arc.
Open GSP File
Now Solve This 2.2. Alternative Proof of Theorem 2.1 -- Open GSP File
The Inscribed Angle Theorem
Inscribed angle -- an angle with its vertex on the circle and whose sides lie along chords of the circle.
Open GSP file
Theorem 2.2: The Inscribed Angle Theorem
The measure of an inscribed angle equals half the measure of the intercepted arc.
Open GSP File Within the GSP File Animate point A anywhere on the arc BAC
PROOF: The proof has been presented in Problem 2.2 (see above).
Corollary 2.1: In any circle, all of the inscribed angles intercepting the same arc are congruent.
Proof: By Theorem 2.2. The proof of the corollary follows by using any two points, P and Q with angles intersecting the same arc
Theorem: 2.3: If the measure of angle A, an inscribed angle subtending chord BC of a circle, then the locus of all points (in the half plane determined by line BC and a point A on the circle) from which the chord BC is seen by the measure of angle A is the arc BAC.
Proof: We need to show that if P is not on the arc BAC but in the half plane as designated, then the measure of angle BPC ≠ the measure of angle BAC. Two cases: P inside the circle and P outside the circle.
In either case the strategy is to construct a point Q on the circle to create an angle BQC which by Corollary 2.1 is congruent to angle BAC. In Case 1, we extend BP to intersect the circle at Q. In Case 2, we select the point Q where BP intersects the circle.
In each case by Corollary 2.1 the measure of the angle at Q is the same as the measure at A. Examine the triangle QPC in Case 1. The angle at P is the exterior angle of triangle QPC and so it is larger than the angle at Q. Therefore the measure of the angle BAC ≠ the measure of angle BPC. In somewhat the same way, examine the triangle PQC in Case 2. Angle BQC is an exterior angle and therefore has greater measure than the angle at P. So the measure of the angle at at A ≠ the measure of the angle at P.
NOW SOLVE THIS 2.3
1. Completion of the proof of Thm 2.3 in Case 2. We have completed that in the previous discussion.
2. Given a segment construct the locus of all the points in the plane as seen by an angle of (a) 45 degrees and (b) 30 degrees.
Investigation: The locus will be the arc of a circle and the given segment will be subtended by the angle of 45 degree. Thus the corresponding arc will have a measure of 90 degrees. Thus the central angle will be 90 degrees. In we had a square constructed on the given segment, its center would be a 90 degree angle and a circle with center at the center of the square and inscribed in the circle would provide the arc of the circle having the desired locus.
Construction: Construct a square on the given segment. Construct circle with center at the center of the square with the given segment on the circle. Select any point on the major arc.
Proof: The angle constructed is the same for all points on the arc (Corollary 2.1). Select a point on the circle at the intersection with a perpendicular from one end of the segment. We have an isosceles right triangle and therefore the angle is 45 degrees.
Investigation: To inscribe an angle of 30 degrees will mean a central angle of 60 degrees so beginning with an equilateral triangle with two vertices on the line segment and the other vertext as the center of the circle will be a good starting point. Now a 30-60-90 right triangle would give three points to determine a circle with an inscribed angle or 30 degrees and the required subtended segment. Any point on the major arc would inscribe a 30 degree angle.
Construction: Construct an equilateral triangle on the given segment. From one end of the segment construct a perpendicular. Extend the otherside of the equilateral triangle to intersect. Construct a circle with center at the non-segment vertex of the equilateral triangle and through the end points of the segment. Any point on the major arc will give the 30 degree inscribed angle.
Proof: The construction produces a 30-60-90 right triangle. The center of the circumcircle is at the midpoint of the hypotenuse. Thus one angle on the major arc that subtends subtends the segment is 30 degrees. By Corollary 2.1 all points on the arc are 30 degrees.
We want to generalize this constuction: Given any segment and any angle, construct a locus of the vertex of the given angle subtending the given segment.
We are given some angle and a segment. In what we have done it the two examples above, a key point was to locate the center of the circle that will contain the inscribed angle.
We know that center will have a central angle AOB that is twice the inscribed angle. Further triangle AOB will be isosceles and so its vertex, point O, will be along the perpendicular bisector of AB.
A perpendicular constructed from one end of the segment will be parallel to the perpendicular bisector with will contain point O.
By alternate interior angles, the give angle is along the perpendicular and half the central angle.
Construct the given segment AB and a construct perpendiculars from its midpoint and from one end.
Copy the given angle with one side along the perpendicular. Extend the other side of the angle to intersect the perpendicular bisector of AB. The center O is located. Contruct the circle center O passing through A and B.
Proof: The construction leads to a right triangle with acute angles alpha and 90 - alpha. The circle containing the locus is determined.
Corollary 2.2: Any angle inscribed in a semicircle is a right angle. Open GSP file.
Proof: By Thm 2.2, any 90 degree angle subtends an arc of 180 degrees and hence the chord is a diameter. Conversely, if the angle ssubtends a diameter, it intercepts an arc of 180 degrees and therefore by Thm 2.2 has measure of 90 degrees.
NOW SOLVE THIS 2.4.
1. Construct a 45 degree angle and use the information from example 2.1 to trisect the angle. In Example 2.1, notice that triangle BOC is an isosceles right triangle. Therefore the angle BOC has been trisected by angle BOD. Bisect angle DBC to get the other two 15 degree trisections.
2. Trisect a 45 degree angle in a different way. One suggestion is given in the following diagram. Construct a square. Consruct a diagonal to create a 45 degree angle. Construct an equilateral triangle inside the square with one vertex in common with a 45 degree angle.
Quadrilaterals Inscribed in a Circle
A polygon is inscribed in a circle if each vertex is on the circle. To say that a polygon is cyclic is the same as that it is inscribed, although a cyclic polygon is assumed to have the vertices named in order around the circle. To say that the circle is circumscribing the polygon is equivalent.
Every triangle is cyclic; it can be inscribed in a circle; it can be circumscribed by a circle.
When is a quadrilateral cyclic?
Precursor to Theorem 2.4: A necessary and sufficient condition for a quadrilateral to be cyclic is that the perpendicular bisectors of the sides are concurrent.
1. If the quadrilateral is cyclic, then the center is equidistant from the endpoints of each side and hence on each perpendicular bisector.
2.If the perpendicular bisectors are concurrent, then the center of concurrency is the center of the circumscribing circle.
In general, this "Precursor" is not very useful since it does not readily lead to a 'test' of whether a quadrilateral is cyclic.
Theorem 2.4: A necessary and sufficient condition for a quadrilateral to be cyclic is that the sum of the measures of a pair of opposite angles is 180 degrees.
Necessary -- if the quadrilateral is inscribed in a circle then the sum of the measures of a pair of opposite sides is 180 degrees.
Proof: Take an inscribed quadrilateral and construct one of its diagonals.
This is what many school mathematics texts consider as the proof of this theorem. It is only the necessity condition proof
The quadrilateral is divided into two triangles. The Angle at B is subtending arc ADC and the angle at D is subtending the arc ABC. The total of the two arcs is 360 degrees and so the the sum of the measures is 180 degrees.
Sufficiency -- If the sum of the measures of opposite angles is 180 degrees, then the triangle is cyclic. That is, given the measures of angle G and angle E sum to 180 degrees.
Assume EFGH is not cyclic and look for a contradiction. Take E to not be on the circle. Construct a point E' by extending side HE to the circle and construct E'F.
By the first part of the theorem, E'FGH has the sum of measures the angles at E' and and G to be 180 degrees. Consider the triangle E'FE. The angle at E is an exterior angle and therefore greater than either of the opposite interior angles. Therefore the angle at E is greater than the angle at E'. This is a contradiction since both of them can not sum with the measure at G to give 180.
In a similar manner, if E was outside of the circle we would build a contradiction and thereby complete the proof.
NOW SOLVE THIS 2.5
Two constructions and explorations to be done with GSP.
1. OPEN GSP File This exploration constructs an inscribed quadrilateral with perpendicular bisectors and then a perpendicular to one of the sides from the point of intersection of the diagonals. You are to conjecture and prove something about the intersection of this latter perpendicular with the opposite side.
2. Draw any quadrilateral ABCD (not necessarily cyclic) and construct the angle bisectors of the four angles. The intersections determine a new quadrilateral XYZW for exploration and conjecture. Open GSP File Run the Animate. A circle through points XY and W has been constructed. As the animation runs, it appears to always go through point Z. Conjecture: the quadrilateral XYZW will be cyclic. Proof?
Discussion/Solutions for Problem Set 2.1