Problem Set 2.2

2.2.1. Triangle ABC is inscribed in a circle. Point D is the center of the inscribed circle. Prove that angle DAE is congruent to angle ADE. (Note:

the 'triangle' in the left hand side of the prove statement should be 'angle')We want to show that triangle EAD is isosceles with the congruent base angles at A and D. It helps to identify the triangle.

Note angle ADE is the exterior angle of triangle ADC so we can express it as the sum and look for substitutions to get the measure of angle DAE)

m(angle ADE) = m(angle CAD) + m(angle DCA)

Since AD is an angle bisector, m(angle CAD) = m(DAB)

m(angle ADE) = m(angle DAB) + m(angle DCA)

Now angle DCA has the same measure as angle ECB because CD is an angle bisector. Both Angle ECB and angle BAE subtend the arc BE so they have the same measure .

Open GSP file.

2.2.2 If a circle with center O is inscribed in a trapezoid whose bases are tangent to the circle at point P and Q (see figure), prove:

a. O, P, and Q are collinear

b. The diameter of the circle equals the height of the trapezoid

c. m(angle COB) = 90 degrees.

Part a seems like a candidate for an indirect proof. Construct a line PO to intersect AB at R at some point other than Q. Show contradiction.

Part b should follow from part a and the fact that PQ must be perpendicular to AB and DC.

Part c. Note OC and OB are angle bisectors.

EXTRA:

Prove line OB is perpendicular to AD and line OA is perpendicular to BC.Prove:

If S and T are respectively the points of tangency on AD and BC, show that ASOQ, DSOP, CPOT, and BQOT are cyclic.Prove:

In any circumscribable polygon, the quadrilateral made up of a vertex, the two adjacent points of tangency of the inscribed circle, and the center of the inscribed circle form a cyclic quadrilateral.

a. Try showing m(angle PAQ) = m(angle PBQ) . Note they each subtend the chord PQ. Since there are congruent circles the arcs the angles subtend have the same measure.

b. If we can show CP = PA, then both are equal to PB from part A so there will be a circle ABC with AC as a diameter and thus triangle ABC is a right triangle.

So, to show CP = PA two approaches seem possible.

The first, knowing AP = PB would be to show PBC is isosceles with base angles at B and C.

The second is to note that APQ is a right triangle because AQ is a diameter. Use that to show PQC is a right triangle with CQ as a diameter. You are almost done.

2.2.4. Two circles which share a common center are called

concentric circles. If segments AB and CD are two chords in the larger circle tangent to the smaller circle, prove that AB = CD.

If you have proved the theorem embedded in part a of Problem 2.1.1, then the proof here can make use of that result.

Otherwise, there are several ways to consider congruent triangles involving radii of the large circle to A, B, C, and D as well as the perpendiculars from O to the points of tangency with the small circle.

Corollaries:

Chords equidistant from the center of a circle are congruent, and conversely.

A perpendicular to a chord from the center of a circle goes to the midpoint of the chord.

2.2.5 Check out the GSP File. We have a circle O with tangents PA and PB. If Q is on arc AB and CD is tangent at Q, prove that

a. m(angle COE) is the same regardless of the position of Q on arc AB.

b. Show the perimeter of triangle PCD is the same regardless of the position of Q on arc AB.

Open the GSP fileand run the animation.Measurement of the angle COD or the perimeter of triangle PCD will confirm that what you want to prove appears to be the case. Yet, no amount of measuring will establish the proof.

If we can show that OC is a bisector of angle BOQ and OD is a bisector of angle AOQ, regardless of where Q is on arc AB, then the angle COD is half the measure of angle BOA. Check out if AOBP is a cyclic quadrilateral.

For part b, look at PC + QD + DP. Can we substitute something?

2.2.7

Open the GSP fileABCD is a quadrilateral with right angles at A and D. Circle O is tangent to the sides AB and AD at B and E, respectively. The diagonal AC contains O. The side CD intersects the circle at P, and PB intersects AC at Q.

a. Find the angles of Triangle CQP

b. Prove that AQ = r, where r is the radius of the circle. That is, AQ ? CO

OB and OE will surely be helpful segments to use. BAEO is a square with side equal to the radius of the circle.

2.2.8

Open GSP file.A circle is tangent to a square ABCD at P with vertices A and D on the circle. The side DC intersects the circle at Q. Prove that

a. Triangle QPA is a right triangle.

b. Segment AP is a bisector of angle QAB

A hint for part a is to examine triangle ADQ.

A hint for part b is to examine the marked angles

2.29 Determine whether it is possible to find a circumscribable, isosceles 'true' trapezoid (parallel sides are not congruent) that has the following properties (three separate questions here). Justify your answers.

Note that if the trapezoid is circumscribable and isosceles, then the sum of the lengths of the bases must be twice the length of the two congruent sides: a + b = 2c.

a. A diagonal bisects an angle of the trapezoid.

b. The trapezoid is cyclic.

Prove or disprove:

A trapezoid is cyclic if and only it is isosceles.Prove or disprove:

Every isosceles triangle is circumscribable.Prove or disprove:

If a trapezoid is circumscribable then it is also cyclicc. The diagonals are perpendicular.

A quadrilateral with perpendicular diagonals is not necessarily a kite or a rhombus.

Prove:

If a circumscribable isosceles has parallel bases a and b and congruent sides c show that there is a rhombus with sides of length c that is circumscribable about the same circle.

2.2.10 a. A circle O is inscribed in a rhombus. ABCD is a quadrilateral whose vertices are the four points of tangency. What kind of quadrilateral does ABCD seem to be? Prove.

Open the GSP file and explore. Or construct a GSP file of your own. It looks, like a rectangle, seems to always be a rectangle as the rhombus is changed, proof?Consider the diagonals of the rhombus and the diagonals of the inscribed quadrilateral. Also, see problem 2.2.2.

b. What is the converse of the proposition proved in part a?

If a rectangle is inscribed in a circle, the circumscribable quadrilateral with points of tangency at the vertices of the rectangle is a rhombus.

2.2.11 A circle is inscribed in a trapezoid ABCD, and MN is the midsegment of the trapezoid. Prove the following.

a. Segment MN contains O.

b. ON = NB and OM = MA

Open the GSP filefor exploration and hints. The trapezoid is not necessarily isosceles.

2.2.12 In a trapezoid ABCD, segment MN is the midsegment and P is a point on MN such that PM = MA and PN = NB.

a. Is it possible to inscribe a circle in ABCD. Justify.

b. Prove that P is the center of the inscribed circle.

We are dealing with the converse of what was shown in Problem 2.2.11.

Consider triangles AMP and BNP.

See GSP file

2.2.13 Prove that the center O1 and O2 of two disjoint circle and the intersection P where two tangents intersect are collinear.

We need to show that centers O1 and O2 are on the angle bisectors.

There are several ways to do this. one is to see that center O1, P, and the two tangent points to circle O1 are a kite and there for the segment O1 to P is along the angle bisector. Then consider the vertical angles at P and argue that center O2 is also on the angle bisector. . .

2.2.14. Two points with only one point in common are called

tangent circles.a. Prove that the common point of the two tangent circles and the centers are collinear. (Hint: Assume the contrary and use the triangle inequality).

b. Prove that the tangent to one of the circles at the point of contact is also tangent to the other circle.

Part a: Assume that Point P is not on the line of centers AB. Then there is a triangle APB with AB being the shortest distance form A to B. The triangle inequality would indicate AP + BP > AB. But the radius is the shortest distance from a center to the circle. Thus the distance from A to P and then from P to B would be less than or equal to any other path. Thus the assumption that P is not on the line of centers must be rejected.

Part b.

See the GSP file. The tangent to circle A at P is a line perpendicular to the line of centers as was shown in part a. But this tangent is also perpendicular to BP at P.

2.2.15. Circles O1 and O2 are tangent at B to each other. A is any other point on the common tangent through B and AP and AQ are tangents to circles O1 and O2 respectively. Prove that AP = AQ.

Use Theorem 2.6

2.2.16 Two circles O1 and O2 are tangent at P. The line AB is tangent to the circles at A and B, respectively. If C is on AB and CP is the common tangent through P, prove that

a. AC = CB

b. m(angle APB) = 90 degrees.

Hint: Use the result of Problem 2.2.15 for part a. For part b construct the circle from the common length of CP = AC = CB

2.2.17. The circles O1 and O2 are tangent at P. A line through P intersects the first circle at A and the second circle at B. Prove that the tangent at A to the first circle is parallel to the tangent at B to the second circle. See GSP File and vary direction of AB through P.

Hint: Examine the line of centers and the two isosceles triangles created by it.

2.2.18 Two congruent circles with centers at D and E intersect at F and A. A third circle with center at B intersect D in B and E in C. (These points are in the same half plane determined by FA.) Prove A, B, and C are collinear.

There are really two cases for this problem. The one depicted here and in the text is for when the radius of the red circle is less than FA.

The second case occurs when the radius is larger than FA but less than twice the radius of the two congruent circles.

A picture is given below.

The situation when FA equals the radius of the red circle will have B and A at the same point.

I will concentrate on the first case. A first hint is to use the Inscribed Angle Theorem to show that angle FAC and angle FABB have the same measure. That does not prove that B is on AC but it is a first step.

2.2.19 The vertices of Triangle EFD are on the sides of Triangle ABC. Construct the circumcircles of Triangle ADE, Triangle DFC, and Triangle EBF. Repeat the construction for differently positioned triangles EFD. (With GSP we can use some animation feature to vary the locations of E, F, and D on their respective sides of ABC).

a. Based on what is observed, make a conjecture about the three circumcircles.

b. Prove your conjecture in part a.

Hint for b. Check out cyclic quadrilaterals.

2.2.20 See GSP File

a. Prove that whenever three circles are tangent to each other (as shown) the three tangents to the points of contact are concurrent.

b. Use the result of part a to show how to construct three noncongruent circles tangent to each other.