Lecture Notes: Introduction to GeometryMontgomery Blair Math Team

10/08/2014

1. Angles and Circles

1.1. Inscribed and Central Angles

Definition 1.1. An inscribed angle of a circle is an angle with its vertex V on the circle,such that the two rays of the angle intersect the circle at two points other than V. Anexample is shown below.

Definition 1.2. A central angle of a circle is an angle with its vertex at the center ofthe circle. We say that the measure of the arc of a circle is equal to the measure of itscorresponding central angle. (For example, if circle O1 has two points A and B such that

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Theorem 1.3 (Inscribed Angle Theorem). Lei O be a circle with inscribed angle AVB,as shown below. Then LAVB — \LAOB (that is, an inscribed angle's measure is half ofthe measure of the corresponding central angle.

Proof. Extend VO to intersect circle O at T. Let LAOT = x and LBOT = y. ThenLAOV = 180° - x and LBOV = 180° - y. Since triangles AOV and BOV are isosceles,LOAV = LOVA = | and LOBV = LOVB = \. Therefore, LAVB = ̂ = \LAOB, asdesired. C

Think about the proof above. It's actually not quite complete. What have we missed?

A andTheorem•MOMMMMHMMMMBH

LAVB

From Theorem 1.3 we conclude that all angles inscribed on the same arc are congruent.This is a very useful fact for our discussion of cyclic quadrilaterals.

By convention, point O refers to the center of circle O.

1.2. Cyclic Quadrilaterals

Definition 1.5. A quadrilateral is cyclic is there exists a circle O such that all fourvertices of the quadrilateral lie on O.

Consider the diagram below.

Theorem 1.3 tells us that the following are equivalent2 (meaning that if you know thatone is true, you know that the others are true):

• ABCD is a cyclic quadrilateral;

• LEAD + LBCD = LADC + LABC - 180°;

• LBAC - LBDC\

• LBAC = LBDC]

• LBAC - LBDC] and

• LBAC = LBDC.

This fact is very useful in many geometry problems. In particular, pay attention toopposite angles being 90°; that is a common way to conclude that a quadrilateral is cyclic.

1.3. Intercepted Arcs

Theorem 1.6. Let AC and BD be chords of a circle that intersect at X. Then LAXD

= \(AD + BC}.

Proof. LAXD = LBXC = 180° - LDXC = LXDC + LXCD = \ BC +\ AD byTheorem 1.3, as desired. D

The proof of the following result is similar:

Theorem 1.7. Let AD and BC be secants of a circle that intersect at X (outside the

circle). Then LCXD = \(CD - AB}.

5You actually need to prove a sort of converse of Theorem 1.3, which we left out.

Think: what happens in the "limiting case" when one of the "secants" is actually atangent?

2. Triangles

2.1. Congruent Triangles

Two triangles are congruent if:

• Their corresponding sides are congruent (SSS);

• Two pairs of corresponding sides and the angles between them are congruent (SAS);

• Two pairs of corresponding angles a,nd any pair of corresponding sides are congruent(ASA if the side is between the angles; otherwise AAS); or

• The triangles have right angles and have a congruent leg and hypotenuse (HL).

If you know that two triangles are congruent, you can use this information to determinethat their corresponding parts are congruent.

2.2. Similar Triangles

Two triangles are similar if:

• Two pairs of angles are congruent (AA); or

• The ratio of the triangles' corresponding sides is constant.

Consider the diagram below, where LABC is a right angle and BD is an altitude toside AC. B

AClearly, &ABC ~ AADB ~ ABDC. Thus, we have

ABAD

BCCD

BD

ACAB

ACBC

DC

AB2 = AD - AC-

AD BD

BC2 = CD • AC] and

BD2 = AD - DC.

3. Angle Chasing

The best way to learn how to angle chase is to do problems. Hence, there is a section onthe problem set devoted to angle chasing. If you happen to get stuck on a question, seeif you can use anything from the list below to help solve the problem.

Sum of angles in a triangleSum of angles in a quadrilateralAngles intercepting arcsAngles which together form a lineAngles around a pointAngles in a right triangleInscribed angles subtending same arcAngle bisectors

Isosceles trianglesEquilateral trianglesVertical angles-Similar trianglesCongruent trianglesParallel lines and anglesExterior anglesPerpendicular lines

4. Power of a Point

Theorem 4.1. Let chords AB and CD of a circle intersect outside the circle at a pointX, as shown below. Then XB • XA = XC • XD.

Proof. Since ABCD is cyclic, we have that LADC = 180° - LABC = LCBX, andwe know that LBXC = LBXC. Thus, &AXD ~ l\CXB. Therefore, ff = f^, soXB • XA = XC - XD, as desired. ' D

The following very important and useful theorems have similar proofs, which are leftas exercises:

Theorem 4.2. Let chords AC and BD of a circle intersect inside the circle at a pointX. ThenXA-XC =

Theorem 4.3. The two tangent segments from a point P outside a circle to the circlehave th& same length. - . ^

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