10.4 - Circles, Arcs, Inscribed Angles, Power of a Point
We work in Euclidean Geometry for the remainder of these notes.
Definition: A minor arc is the intersection of a circle with a central angle and its interior. A semicircle is the intersection of a circle with a closed half-plane whose boundary line passes through its center. A major arc is the intersection of a circle with a central angle and its exterior (that is, the complement of a minor arc, plus its endpoints).
Notation: If the endpoints of an arc are A and B, and C is any other point of the arc (which must be used in order to uniquely identify the arc) then we denote such an arc by qACB .
Measures of Arcs: We define the measure of an arc as follows:
Minor Arc Semicircle Major Arc mqACB = µ(pAOB) mqACB = 180 m=qACB 360 - µ(pAOB)
A A A
C
O O C O C
B B B
q q Theorem (Additivity of Arc Measure): Suppose arcs A1 = APB and A2 =BQC are any two
arcs of a circle with center O having just one point B in common, and such that their union A1 c q A2 = ABC is also an arc. Then m(A1 c A2) = mA1 + mA2.
The proof is straightforward and tedious, falling into cases depending on whether qABC is a minor arc or a semicircle, or a major arc. It is mainly about algebra and offers no new insights. We’ll leave it to others who have boundless energy and time, and who can wring no more entertainment out of “24.”
Lemma: If pABC is an inscribed angle of a circle O and the center of the circle lies on one of its 1 sides, then μ()∠=ABC mACp . 2
~ WLOG suppose O lies on side BC , and construct radius OA . Note that since they are both radii, OA = OB. Thus, triangle ªAOB is isosceles, with µ(pABO) = µ(pBAO). By the Exterior Angle Theorem, µ(pAOC) = µ(pABO) + µ(pBAO )= 2 µ(pABC). Thus, 11 μμ()()∠=∠=ABC AOC mACp . 22
A
B O C
Theorem (Inscribed Angle Theorem): The measure of an inscribed angle equals one-half the measure of the arc it intercepts.
~ Let pABC be an inscribed angle of a circle O. There are three cases. Case 1: O lies on BC . This is the case of the above Lemma, which we have already proved.
Case 2: O is in the interior of pABC. Let BD be a diameter. Since O is in the interior, we μ()()()∠ABC=∠μμ ABD +∠ DBC have angle additivity. Using the Lemma, 11 1 =+=mADp mDCp mACp 22 2
A
O D B
C JJJG JJJG JJJG JJJG JJJG JJJG Case 3: O is in the exterior of ABC. Thus either BABCBD** or BCBABD** (the third JJJG JJJG JJJG p JJJG JJJG JJJG option, BABDBC** , implies O is interior to pABC). WLOG, suppose BABCBD** . Once again, by angle additivity, the Lemma, and additivity of arcs, we have .
μ()∠=∠−∠ABC μμ()() ABD CBD 111 =−=mACDq mCDp mACp 222 A
C D B O
Corollary: An angle inscribed in a semicircle is a right angle.
Corollary: If a quadrilateral QABCD is inscribed in a circle, opposite angles are supplementary.
Corollary: An angle whose vertex lies inside a circle and is formed by intersecting chords of the 1 circle (intercepting arcs of measure x and y) has measureθ = ()x + y . 2
D
y B A x E C
~ Note µ(pABC) + µ(pBAC) + µ(pACB) =180. Also, mECp +++= mDBp x y 360 . So, 111 mECp +++= mDBp ()180 x y . Now 222 11 μμ()∠=ABC mECp ; ∠= ACB ) mDBp , so 22 1 μμ()()()180()()()∠ABC +∠ ACB + x += y =∠ μμμ ABC +∠ ACB +∠ BAC or 2 1 μ()()∠=+BAC x y . 2 Corollary: An angle whose vertex is exterior to a circle and is formed by intersecting secants of the circle (intercepting arcs of measure x and y) has measure 1 θ =−x y . 2 B x
A y C
~ First note that 1111 mDBp +++= mECp x y360; mDBp + mECp ++= x y 180 Adding and 2222 1 111111 subtracting a y we get: mDBmECxyyyp +p +++−=180 , 2 222222 111111 11 or mDBp ++ y mECp ++−= y x y 180 . But, μ()∠=ABC mECp + y , 222222 22
1111111 11mDp B+ym+ECp +yx−=y∠mABCm+∠ACB+−x()1y8=0 1 and μ()∠=ACB mDBp + y . So, 22222 2 μμ(∠+∠+−=ABC ) ( ACB ) ( x y ) 180 , and since 22 2 1 μ()()()180∠+∠+∠=ABCμμ ACB BAC , we must have μ()()∠BAC=− x y . 2
B
x
C D y A E Corollary: An angle formed by a chord and tangent of a circle, with its vertex at the point of 1 tangency and intercepting an arc of measure x on that circle, has measure θ = x . 2
B
x
C
A
~ Make a diameter AD , and use the fact that pDAB cuts of the arc 1 DpB and so has measure mDBp , and mDBp += x 180 . The rest follows from the fact that 2 µ(pBAC) = 90 - µ(pDAB).
D
B O
x
A C Theorem (Two Chord Theorem): When two chords of a circle intersect, the product of the lengths of the segments formed on one chord equals that on the other chord. That is, referring to the figure, AP @ PB = CP @ PD.
~ Note that inscribed angles pACD and pDBA (1 and 2 in the figure) both cut off the arc ApD and so are congruent. Also, pAPC and pBPC are vertical angles and so are congruent. Thus, by the AA Similarity theorem, ªACP is similar to ªDBP. This immediately gives APCP = , and clearing fractions gives the result. PD PB
C 1
A 3 P B 4 2
D HJJG HJJG Theorem (Secant-Tangent Theorem): If a segment PA and tangent PC meet the circle at the respective points A, B, and C (point of contact) then, referring to the figure, PC2 = PA@ PB.
~ Construct segments AC and BC . Since pCBA cuts of arcpAC it 1 has measure mACp . B 2 By Corollary E above, so does pACP. Now since ªPBC and ªPCA share two congruent angles they are similar under this A PC PB correspondence. Then = , and clearing fractions gives the PA PC result. P C
HJJG HJJG Corollary (Two-Secant Theorem): If two secants PA and PC of a circle meet the circle at A, B, C, and D, respectively, then (referring to the figure) PA @ PB = PC @ PD
~ Draw a tangent from P to point E, and apply the Secant-Tangent Theorem to both secants:
PA @ PB = PE2 = PC @ PD B
A
P
C D E
Note that combining the last three results proves that, if P is any point not on a circle, and if P lies on two different secant lines l and m that intersect the circle at points A & B and at points C & D respectively, then (PA)(PB) = (PC)(PD). This number, which is the same for any secant line containing P, is called the power of the point PHJJ withG respect to the circle. Moreover, if T is a point on the circle and P is external to the circle, PT is a tangent line, and the (PT)2 is also equal to the power of the point P relative to the circle.