Cyclic Quadrilaterals

Andrew Tung March 2021

1 Introduction

Deﬁnition A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle. B

Four points A, B, C, and D are called concylic if they all lie on the same circle (i.e. quadrilateral ABCD is cyclic). The main theorem regarding cyclic quadrilaterals (with regard to angle chasing) is the following.

Proposition 1 If ABCD is a convex quadrilateral, the following are equivalent.

1. ABCD is cyclic

2. ∠DCA = ∠DBA

3. ∠CDA = 180 − ∠CBA The proof of this is by the inscribed angle theorem, and is not hard1. Note that Propo- sition1 can be used to prove a quadrilateral is cyclic, or to get some angle information from

1Left as an exercise to the reader.

1 a known cyclic quadrilateral. Both directs are important.

Cyclic quadrilaterals also give similar triangles: Proposition 2 If ABCD is a cyclic quadrilateral whose diagonals intersect at P , then 4PBA ∼ 4PCD and 4PBC ∼ 4P AD. B

P C

Proof We prove 4PBC ∼ 4P AD; the other case is symmetric. Using Proposition1, ∠DBC = ∠DAC. So ∠PBC = ∠DAP . Also ∠BPC = ∠AP D because they are vertical angles. Hence by AA similarity, 4PBC ∼ 4P AD. Note that Proposition2 does not imply AB||CD, since that similarity would go the “other way” (i.e. 4PBA ∼ 4PDC). Also note that the side ratios resulting from Proposi- tion2 imply Power of a Point: BP · PD = AP · PC.

Some other computational tools relating to cyclic quadrilaterals include: Proposition 3 (Ptolemy’s Theorem) In a cyclic quadrilateral ABCD, we have

(AB)(CD) + (BC)(DA) = (AC)(BD) The proof of Ptolemy’s Theorem is either by trig (which is slightly messy), or by con- structing an isogonal conjugate of a diagonal, which leads to some similar triangles. Proposition 4 (Brahmagupta’s Formula) If a cyclic quadrilateral has side lengths a, b, c, d, and if a + b + c + d s = 2 then the area of the quadrilateral is

p(s − a)(s − b)(s − c)(s − d) The proof of Brahmagupta’s Formula is by trig or by repeated applications of Heron’s formula. Both are messy (and are omitted).

2 2 Examples

2.1 Using cyclic quads for angle chasing These ﬁrst two problems illustrate one of the basic uses of cyclic quadrilaterals: angle chasing.

Problem (2006 AIME II 12) Equilateral 4ABC is inscribed in a circle of radius 2. Extend AB through B to point D so that AD = 13, and extend AC through C to point E so that AE = 11. Through D, draw a line l1 parallel to AE, and through E, draw a line l2 parallel to AD. Let F be the intersection of l1 and l2. Let G be the point on the circle that is collinear with A and√ F and distinct from A. Given that the area of 4CBG p q can be expressed in the form r , where p, q, and r are positive integers, p and r are relatively prime, and q is not divisible by the square of any prime, ﬁnd p + q + r.

C B

The key claim is that 4BGC ∼ 4FDA To prove this, we use the cyclic quadrilateral ABGC. Note ∠BCG = ∠BAG = ∠F AD since ABGC is cyclic. Similarly, ∠CBG = ∠CAG = ∠EAF = ∠AF D, using the fact that

3 ADF E is a parallelogram.

So to find the area of 4CBG, we first find the area of 4FDA, and then multiply√ by BC2 143 3 AF 2 . Note ∠FDA = ∠BGC = 180 − ∠BAC = 120, so the area of 4CBG is 4 . Using Law of Cosines, AF 2 = 433. Hence our answer is √ √ 143 3 4 143 3 = 4 433 433

Problem (2018 AMC 12A 20) Triangle ABC is an isosceles right triangle with AB = AC = 3. Let M be the midpoint of hypotenuse BC. Points I and E lie on sides AC

and AB, respectively, so that AI > AE and AIME is a cyclic√ quadrilateral. Given that a− b triangle EMI has area 2, the length CI can be written as c , where a, b, and c are positive integers and b is not divisible by the square of any prime. What is the value of a + b + c?

A X I C

In this case the cyclic quadrilateral is given: AIME. We note that ∠MAI = ∠MAB = 45, since M is the midpoint and ABC is isosceles. So ∠MEI = ∠MAI = 45, and 4EMI is isosceles. Hence MI = 2.

Let the foot of the perpendicular from M to AC be X. Then MX = 3 , and since √ √ 2 7 3− 7 MI = 2, XI = 2 by the Pythagorean Theorem. So CI = 2 .

4 2.2 Using cyclic quads for length ratios These next two problems illustrate the use of cyclic quadrilaterals for ﬁnding ratios of lengths, using the similar triangles in Proposition2.

Problem (2002 AIME I 13) In triangle ABC the medians AD and CE have lengths 18 and 27, respectively, and AB = 24.√ Extend CE to intersect the circumcircle of ABC at F . The area of triangle AF B is m n, where m and n are positive integers and n is not divisible by the square of any prime. Find m + n.

B A E

The midpoints in this problem suggest area ratios. Note that [4AGE] is known, since AG = 12, AE = 12, and GE = 9. Hence we relate [4AF B] to [4AGE].

1 GE 1 Note that [4AGE] = 3 [4ACE] since CE = 3 . Also note [4AF B] = 2[4EFB] since E is the midpoint of AB. But 4ACE ∼ 4FEB since ACBF is cyclic. The ratio of similarity CE 27 is just EB = 12 .

5 √ 27 55 A simple computation reveals [4AGE] = 4 . So as outlined above [4AF B] = 2[4EFB] 122 = 2 [4ACE] 27 42 = 2 (3)[4AEG] 9 √ ! 32 27 55 = 27 4 √ = 8 55

Problem (2021 AIME I 11) Let ABCD be a cyclic quadrilateral with AB = 4,BC = 5,CD = 6, and DA = 7. Let A1 and C1 be the feet of the perpendiculars from A and C, respectively, to line BD, and let B1 and D1 be the feet of the perpendiculars from B m and D, respectively, to line AC. The perimeter of A1B1C1D1 is n , where m and n are relatively prime positive integers. Find m + n.

B1 X C A

D We focus on B and C for now, since the other vertices are symmetric. Note that ∠BB1C = ∠BC1C = 90, so quadrilateral BB1C1C is cyclic. So 4B1XC1 ∼ BXC, so B X B C 1 = 1 1 BX BC 6 B1X But note BX = cos ∠AXB. Extending this to the other sides, we note that each side of A1B1C1D1 is simply cos ∠AXB times the corresponding side of ABCD. So it suﬃces to ﬁnd cos ∠AXB. √ This is not hard. By Brahmagupta’s, [ABCD] = 2 210, and by Ptolemy’s Theorem, the product of the diagonals is 59. So √ 1 [ABCD] = 2 210 = (59)(sin AXB) 2 ∠ √ 4 210 11 11 and sin ∠AXB = 59 . Thus cos ∠AXB = 59 , so the perimeter of A1B1C1D1 is 59 (4 + 242 5 + 6 + 7) = 59 .

2.3 Finding cyclic quads The previous problem shows that sometimes, important cyclic quadrilaterals are not immediately obvious. The following two problems continue this theme.

Problem (1991 AIME 12) Rhombus P QRS is inscribed in rectangle ABCD so that vertices P , Q, R, and S are interior points on sides AB, BC, CD, and DA, respectively. It is given that PB = 15, BQ = 20, PR = 30, and QS = 40. Let m/n, in lowest terms, denote the perimeter of ABCD. Find m + n.

AB P

R CD

First note that ∠PXQ = 90, since P QRS is a rhombus. Since ∠PBQ = 90, P BQX is cyclic. Also note that X is the center of the rectangle ABCD, since the height of X is half of BC (using segment PR) and similarly for the horizontal direction.

Using the cyclic quadrilateral P BQX, ∠XPQ = ∠XBQ. Since 4PXQ and 4BCD are both right and share an angle, 4PXQ ∼ 4BCD.

7 Using Ptolemy’s on P BQX, we ﬁnd BX = 24, so BD = 48 and the similarity ratio is 25 28 672 48 . So the perimeter of ABCD is 2(BC + CD) = 2 25 (PX + XQ) = 5 .

Problem (2011 AIME II 10) A circle with center O has radius 25. Chord AB of length 30 and chord CD of length 14 intersect at point P . The distance between the midpoints 2 m of the two chords is 12. The quantity OP can be represented as n , where m and n are relatively prime positive integers. Find the remainder when m + n is divided by 1000.

AB E P F

D O

Let E and F be the midpoints of AB and CD, respectively. The key is to construct OE and OF . Note that since E and F are midpoints, ∠PEO = ∠PFO = 90. So EPFO is cyclic. Observe that since ∠PEO = 90, OP is the diameter of the circumcircle of OEF . So it suffices to find the circumradius of OEF . But it is not hard to find OE = 20 and OF = 24, and it is given EF = 12. So 2 √ (20)(12)(24) [4OEF ] = 2 14 and R = √ = √45 . Thus OP 2 = √90 = 4050 . 4(2 14) 14 14 7

8 3 Takeaways

Sometimes cyclic quadrilaterals are immediately obvious. Typically cyclic quads are not given explicitly in the problem statement, but instead appear because 4 points (or more) lie on a circle by construction. In these cases, it is often helpful to connect all the sides/diagonals of the quadrilateral (if not already drawn), and consider Propositions1 and2. The best indicator that Proposition1 might be helpful is the presence of parallel lines, isosceles triangles, or lots of information given in terms of angles (e.g. angle bisectors). Proposition2 is best used when there are side/area ratios involved, although sometimes it is just a ﬁnishing step. Finding cyclic quadrilaterals, however, can be tricky, especially when the obvious one doesn’t seem very helpful. One of the biggest clues that there might be cyclic quads is if there are some weird angle conditions, for example if two seemingly random angles are the same, or if they are complementary. It might be intuitively clear whether those angles “face the same way” or “face each other,” which would be setups for Proposition1. Another clue is the presence of many right angles, most prominently used in the orthocenter conﬁguration. In these cases it helps to mark the right angles, so that it becomes more visually clear.

9 4 Problems

Here are some problems, roughly ordered in diﬃculty. Note that the last few are very hard.

Problem Prove that a cyclic trapezoid must be isosceles.

◦ Problem (2020 SMT Geometry 2) Let 4ABC be a right triangle with ∠ABC = 90 . Let the circle with diameter BC intersect AC at D. Let the tangent to this circle at D AE intersect AB at E. What is the value of BE ?

Problem (2019 SMT Geometry 6) Let the altitude of 4ABC from A intersect the circumcircle of 4ABC at D. Let E be a point on line AD such that E 6= A and AD = DE. If AB = 13, BC = 14, and AC = 15, what is the area of quadrilateral BDCE? A

B C

Problem (Miquel Point) Let D,E,F be points on sides BC, CA, and AB of 4ABC, respectively. Prove that the circumcircles of 4AEF , 4BFD, and 4CDE concur. (This point is known as a Miquel Point.)

10 Problem (EGMO Problem 1.38) In cyclic quadrilateral ABCD, let I1 and I2 denote the incenters of 4ABC and 4DBC, respectively. Prove that I1I2BC is cyclic.

Problem (2009 AIME I 15) In triangle ABC, AB = 10, BC = 14, and CA = 16. Let D be a point in the interior of BC. Let points IB and IC denote the incenters of triangles ABD and ACD, respectively. The circumcircles of triangles BIBD and CIC D meet at distinct points√ P and D. The maximum possible area of 4BPC can be expressed in the form a − b c, where a, b, and c are positive integers and c is not divisible by the square of any prime. Find a + b + c.

Problem (2019 SMT Geometry 9) Let ABCD be a quadrilateral with ∠ABC = ◦ ∠CDA = 45 , AB = 7, and BD = 25. If AC is perpendicular to CD, compute the length of BC. (Warning: Hard)

Problem (2014 AIME I 15) In 4ABC, AB = 3, BC = 4, and CA = 5. Circle ω intersects AB at E and B, BC√ at B and D, and AC at F and G. Given that EF = DF DG 3 a b and EG = 4 , length DE = c , where a and c are relatively prime positive integers, and b is a positive integer not divisible by the square of any prime. Find a+b+c. (Warning: Hard)

11 4.1 Hints 1. (a) Let one of the base angles be θ. Then angle chase. (b) A trapezoid is isosceles if the base angles are the same.

2. (a) Let O be the midpoint of BC. (b) Examine quadrilateral EBOD. (c) Angle chase. AE (d) BE is a special number. (e) Prove that 4AED is isosceles.

3. (a) BDCE is kind of weird. It might help to use area ratios. (b)[ BDCE] = [ABDC]. Why? 1 (c)[ ABDC] = 2 (AD)(BC). (d) Use Power of a Point/similar triangles.

4. (a) Let the circumcircles of 4AEF and 4BFD intersect at P . It suﬃces to show that the circumcircle of 4CDE passes through P . (b) To prove that the circumcircle of 4CDE passes through P , prove CDPE is cyclic. (c) Prove ∠DCE = 180 − ∠DPE. (d) Angle chase using the other cyclic quads.

5. (a) A helpful lemma is that if I is the incenter of 4ABC, then ∠BIC = 90 + ∠A/2. Prove it.

(b) Prove that ∠BI1C = ∠BI2C. (c) Use the fact that ∠BAC = ∠BDC. 6. (a) Use the lemma from the previous problem.

(b) Use the cyclic quadrilaterals BIBDP and CIC DP . (c) The angle ∠BPC is constant. Why? (d) The area [4BPC] is minimized when PB = PC. (e) We can actually ﬁnd ∠BPC.

7. (a) Since ∠CDA = ∠CBA, AD is tangent to the circumcircle of 4ABC. Why? (b) Extend DC past C until it hits the circumcircle of 4ABC. Call this point E. (c) By cyclic quads, ∠CEA = ∠CBA = 45. So 4ECA is isosceles right. (d) ∠EBA = 90. (e) Extend AB past A until it hits the circumcircle of 4CDA. Call this point F . (f) Prove that 4EBA ∼= 4AF D.

12 (g) Using the side lengths given, ﬁnd BF . (h) Prove that ∠BCF = 90. 8. (a) Note 4GDE ∼ 4BAC. Why? (b) Consider cyclic quadrilateral EF GD. (c) Using the cyclic quad and the similar triangles, prove ∠EAF = ∠AF E. So 4AEF is isosceles. (d) Similarly prove ∠DFC = ∠DCF . (e) Next consider cyclic quadrilateral BEFD. (f) Prove BF is an angle bisector. (g) Find AF and FC. (h) Using the isosceles triangles, ﬁnd AE and DC. (i) Find EB and BD.