cyclic quadrilaterals

Basically, a cyclic quadrilateral is a quadrilateral inscribed in a circle. In other words, quadrilateral ABCDABCD is cyclic iff there exists a circle passing through all four of its vertices, or equivalently, the circumcircle of any three of the points passes through the fourth point. The following figure shows four points A,B,C,DA, B, C, D lying on a circle centered at OO.

Image link- http://s27.postimg.org/c18kjyjo3/Untitled.png Image link- http://s27.postimg.org/c18kjyjo3/Untitled.png

Now, we explore some interesting properties about cyclic quadrilaterals. The following property holds for any three points on a circle.

The angle subtended by an arc at the circumference is half the angle subtended at the center.

Okay, that looks a bit contrived. Let's be a little bit elaborate. Consider two points A,BA, B lying on a circle with center OO. Let CC be any other point on the circle as shown in the figure below. Then, the angle subtended by AB^\widehat{AB} at the center is defined as AOB,\angle AOB, and the angle subtended by AB^\widehat{AB} at the circumference is ACB\angle ACB. We're going to prove that AOB=2ACB.\angle AOB = 2 \angle ACB.

Image link- http://s21.postimg.org/bkzhdoyjr/Untitled.png Image link- http://s21.postimg.org/bkzhdoyjr/Untitled.png

Okay, so let's extend COCO to any arbitrary point XX.

image link- http://s30.postimg.org/96woykrgh/Untitled.png image link- http://s30.postimg.org/96woykrgh/Untitled.png

What can we say about OAC\triangle OAC? It's isoceles! To be a bit more detailed, since AA and CC lie on a circle centered at O,O, so we must have OA=OC.OA=OC. From isoceles OAC,\triangle OAC, this implies OAC=OCA.\angle OAC= \angle OCA. Now, we use the fact that the angles of a triangle sum up to 180.180^{\circ}. OAC+OCA+AOC=180    AOC=1802OCA.\angle OAC + \angle OCA + \angle AOC = 180^{\circ} \implies \angle AOC = 180^{\circ} - 2\angle OCA. Now look at straight line COX.COX. We have COA+AOX=180\angle COA + \angle AOX = 180^{\circ} (why?), so AOX=180AOC=2OCA.\angle AOX = 180^{\circ} - \angle AOC = 2 \angle OCA. We apply similar arguments to OAB\triangle OAB to deduce that XOB=2OCB.\angle XOB = 2\angle OCB. Now, notice that AOB\angle AOB is equal to sum of angles AOX+XOB,\angle AOX + \angle XOB, and ACB\angle ACB is equal to the sum of angles ACO\angle ACO and BCO.\angle BCO. Isn't the result obvious now? Indeed, AOB=AOX+XOB=2(ACO+OCB)=2ACB.\angle AOB = \angle AOX + \angle XOB = 2 ( \angle ACO + \angle OCB) = 2 \angle ACB. Phew! We're done...!

Wait, not yet. Draw a diagram carefully - might we be missing something? Indeed, our proof is only half complete. We fell victims to one of the most notorious yet unnoticeable issues in solving geometry problems: the configuration issue. Is AOB\angle AOB always equal to the sum of AOX\angle AOX and XOB\angle XOB? The answer is no. Look at the following picture.

image link- http://s29.postimg.org/vu0f0uarr/Untitled.png image link- http://s29.postimg.org/vu0f0uarr/Untitled.png

As you can see, AOB\angle AOB is actually equal to the difference between BOX\angle BOX and AOX.\angle AOX. If we let CC slide to the other side of O,O, AOB\angle AOB becomes the difference of AOX\angle AOX and BOX\angle BOX. Fortunately, these cases are handled similarly. We also have that ACB=OCBOCA.\angle ACB = \angle OCB - \angle OCA. It is up to the reader to complete the proof; it's not much different from what we've done already.


Now that we know this interesting fact, let's see some corollaries.

  • If ABAB is a diameter of a circle and CC any other point on that circle, ACB=90\angle ACB = 90^{\circ}.

image link- http://s29.postimg.org/5spyb9ipz/Untitled.png image link- http://s29.postimg.org/5spyb9ipz/Untitled.png

The proof almost immediately follows from what we have proven just before. Suppose OO is the center of the circle. What does ABAB being a diameter of a circle mean? Of course, it mens its midpoint is the center of the circle. Then, AOB=180\angle AOB = 180^{\circ} since A,O,BA, O, B are collinear. We've just proven that ACB=AOB2.\angle ACB = \dfrac{\angle AOB}{2}. Is the conclusion immediate now?

P.S: There is another nice proof of this fact without using any prior knowledge. Hint: reflect CC over ABAB to get a point C.C'. Can you prove CACBCAC'B is a rectangle?

P.P.S: The converse of this statement holds too. If CC is any point such that ACB=90,\angle ACB = 90^{\circ}, CC lies on the circle with diameter ABAB. Can you prove this? If no, look below. The proof to the converse of the property below does the job for this one too.

  • The angles inscribed in the same arc are equal. In other words, consider two points A,BA, B on a circle and any other point CC on that circle not lying between AA and BB. As CC varies along the circumference of the circle, ACB\angle ACB remains constant. Alternatively, if CC and DD are two points on that circle, neither of them lying between AA and B,B, ACB=ADB.\angle ACB = \angle ADB.

image link- http://s28.postimg.org/cjzyz93ot/Untitled.png image link- http://s28.postimg.org/cjzyz93ot/Untitled.png

This proof is also immediate. Let OO be the center of the circle. We have already proven that AOB=2ACB\angle AOB = 2 \angle ACB and AOB=2ADB\angle AOB = 2 \angle ADB. Combining them, the result immediately follows.

Okay, so this is an important property of cyclic quadrilaterals. If ABCDABCD is a cyclic quadrilateral with its points lying on that order, ACB=ADB\angle ACB = \angle ADB (cyclic variations hold). Does the converse also hold true? If quadrilateral ABCDABCD with its points lying in that order satisfies ACB=ADB,\angle ACB = \angle ADB, must it be cyclic?

Yes, it must be cyclic. Let's see why. Suppose it weren't cyclic. Then, the circle passing through A,D,BA, D, B (i.e. the circumcircle of ADB\triangle ADB) would intersect line ACAC at points AA and E,E, where EC.E \neq C. We know that AEB=ADB.\angle AEB = \angle ADB. Hence, AEB=ACB.\angle AEB = \angle ACB. Can we conclude E=CE = C from this? Indeed, we can. For any angle θ\theta, two fixed points A,BA, B and a line \ell passing through AA, there exists exactly one point PAP \neq A on \ell such that APB=θ\angle APB = \theta . Try to visualize this as a point moving on the line. As the point approaches A,A, APB\angle APB increases, and as it moves away, the angle decreases. Hence, EE must be the same point as CC.

image link- http://s4.postimg.org/exr55kr9p/Untitled.png image link- http://s4.postimg.org/exr55kr9p/Untitled.png

  • The opposite angles of a cyclic quadrilateral sum up to 180180^{\circ}. If ABCDABCD is a cyclic quadrilateral, ABC+CDA=DAB+BCD=180.\angle ABC + \angle CDA = \angle DAB + \angle BCD = 180^{\circ}.

image link- http://s28.postimg.org/w52e1hl2l/Untitled.png image link- http://s28.postimg.org/w52e1hl2l/Untitled.png

This proof is also pretty simple. Let's use what we proved earlier. Notice that CAB=CDB\angle CAB = \angle CDB and ACB=ADB\angle ACB = \angle ADB since they are angles in the same arc. Hence, BDC+ABC=BDC+ADB+ABC=ABC+BAC+BCA.\angle BDC + \angle ABC = \angle BDC + \angle ADB + \angle ABC = \angle ABC + \angle BAC + \angle BCA. What is the sum of the angles CAB+ABC+BCA\angle CAB + \angle ABC + \angle BCA? Well, the sum of the angles of a triangle is 180180^{\circ}, so CAB+ABC+BCA=180.\angle CAB + \angle ABC + \angle BCA = 180^{\circ}. Combining them yields the result.

Does the converse hold? Indeed it does! The same argument as in the last paragraph does the job.


Now, let us look at an example problem.

Problem (level: trivial): In ABC,\triangle ABC, MM is the midpoint and E,FE, F are the feet of perpendiculars from B,CB, C on AC,ABAC, AB respectively. Prove that ME=MF.ME = MF.

image link- http://s2.postimg.org/e160mccfd/Untitled.png image link- http://s2.postimg.org/e160mccfd/Untitled.png

At first, one might be tempted to use cosine rule. Indeed, there's nothing wrong with using cosine rule here. Given the side lengths of ABC,\triangle ABC, we already know the values of CM,CECM, CE and cosC,\cos C, so we can easily compute MEME and show that ME=MC.ME=MC. However, let's not do it the hard way. Let's see what we're given to prove.

We wish to prove ME=MF.ME=MF. There are several approaches which look promising. We can angle chase and try to prove that MEF=MFE.\angle MEF = \angle MFE. However, apparently there's no "easy" path that seems to lead to the values of MEF\angle MEF and MFE.\angle MFE. Instead, let's prove that EE and FF lie on a circle centered at MM. Indeed, this is true: since BEC=90,\angle BEC = 90^{\circ}, EE lies on the circle with diameter BCBC. Since BFC=90,\angle BFC = 90^{\circ}, FF also lies on the circle with diameter BCBC. What is the center of the circle with diameter BCBC? Of course it's MM! So, we're done. \blacksquare Notice that in fact ME=MF=MB=MC.ME=MF=MB=MC.

Now, let's look at another trivial problem.

Problem: Prove that in ABC,\triangle ABC , the internal angle bisector of BAC\angle BAC and perpendicular bisector of BCBC meet on the circumcircle of ABC\triangle ABC.

image link- http://s1.postimg.org/hkq6cp1cv/Untitled.png image link- http://s1.postimg.org/hkq6cp1cv/Untitled.png

Basically, the internal angle bisector of BAC\angle BAC is a line which divides BAC\angle BAC into two equal parts. In other words, if JJ is any point lying on the angle bisector of BAC,\angle BAC, BAJ=JAC\angle BAJ = \angle JAC. The perpendicular bisector of BC,BC, of course, is a line perpendicular to BCBC passing through the midpoint MM of BCBC. In fact, if JJ is any point lying on the perpendicular bisector of BC,BC, it is well known that BJ=CJBJ=CJ (why? hint: BJMCJM\triangle BJM \cong \triangle CJM).

Now let's see what we're given to prove. Suppose the internal angle bisector of BAC\angle BAC meets the perpendicular bisector of BCBC at DD. We are to prove DD lies on the circumcircle of ABC,\triangle ABC, or that quadrilateral ABDCABDC is cyclic. Okay, this looks messy. Let's try something else.

If we can show that the point where the internal angle bisector of BAC\angle BAC intersects the circumcircle again (apart from AA) also lies on the perpendicular bisector of BCBC, we will be done. This idea seems a lot more feasible.

So, suppose the internal angle bisector of BAC\angle BAC meets the circumcircle of ABC\triangle ABC again at DD. We need to prove that DB=DC,DB = DC, or equivalently, DBC=DCB\angle DBC = \angle DCB. We are given that BAD=DAC,\angle BAD = \angle DAC, and we need to prove that DBC=DCB.\angle DBC = \angle DCB. But this is obvious - being angles in the same arc, BAD=BCD\angle BAD = \angle BCD and DAC=DBC.\angle DAC = \angle DBC. So, we're done. \blacksquare

Let's now look at another trivial problem.

Problem: Let HH be the orthocenter of ABC\triangle ABC and AA' the diametrically opposite point of AA on the circumcircle of ABC\triangle ABC. Prove that HAHA' passes through the midpoint of BCBC.

image link- http://s9.postimg.org/aned2p233/Untitled.png image link- http://s9.postimg.org/aned2p233/Untitled.png

Basically, HH is the point of intersection of the altitudes from the points to the opposite sides. In other words, HH is the unique point such that BHAC,AHBC,CHABBH \perp AC, AH \perp BC, CH \perp AB. Also, AA' is the unique point such that AAAA' is a diameter of the circumcircle of ABC\triangle ABC. What do we get from this? Well, from what we've seen before, ABABA'B \perp AB and ACACA'C \perp AC. Can we combine them somehow? Lines ABA'B and CHCH are both perpendicular to ABAB. What does this mean? Well, they are parallel! So, ABCHA'B \parallel CH. Similarly, we can show that ACBHA'C \parallel BH. What can we say about quadrilateral BACHBA'CH? It has two parallel pairs of sides, so of course it's a parallelogram. Now, we're almost done. We need to prove that the diagonal HAHA' cuts diagonal BCBC at its midpoint, i.e. bisects BCBC. However, this is well known - the diagonals of a parallelogram bisect each other (why? hint: if HAHA' intersects BCBC at M,M, BHMAMC\triangle BHM \cong \triangle A'MC), so HAHA' passes through the midpoint of BCBC. Notice that as a consequence, the midpoint of BCBC is in fact the midpoint of HA.HA'.

Now, let's try a slightly harder problem.

Problem (IMO Shortlist 2010 G1): Let ABCABC be an acute triangle with D,E,FD, E, F the feet of the altitudes lying on BC,CA,ABBC, CA, AB respectively. One of the intersection points of the line EFEF and the circumcircle is P.P. The lines BPBP and DFDF meet at point Q.Q. Prove that AP=AQ.AP = AQ.

There are several configurations for this problem. We shall work with the following one. Let's start by drawing a diagram.

image link-http://s17.postimg.org/g21of10y7/Untitled.png image link-http://s17.postimg.org/g21of10y7/Untitled.png

Let's see what we wish to prove. We need to prove that APQ\triangle APQ is isoceles with AP=AQAP=AQ. Let's try angle chasing. To show AP=AQ,AP=AQ, it shall be enough to show APQ=AQP\angle APQ = \angle AQP. But what is APQ\angle APQ? Well, let's look at the picture: APQ\angle APQ is equal to APB,\angle APB, which is equal to ACB\angle ACB by angles in the same arc. Hence, we need to prove that AQP=ACB\angle AQP = \angle ACB. Let's see how many other angles are equal to ACB\angle ACB. Look back at the first example- E,FE, F lie on a circle with diameter BC,BC, so BCEFBCEF is cyclic. What can we say about ACB\angle ACB about this? From the fact that the opposite angles of a cyclic quadrilateral sum up to 180,180^{\circ}, ACB+BFE=180.\angle ACB + \angle BFE = 180^{\circ}. Now, what does this force ACB\angle ACB to be equal to? Well, notice that AFE+BFE=180.\angle AFE + \angle BFE = 180^{\circ}. This forces AFE=ACB\angle AFE = \angle ACB. Now, we need to prove that AQP=AFE\angle AQP = \angle AFE . Notice that AFE=AFP\angle AFE = \angle AFP. Now, what does AQP=AFP\angle AQP = \angle AFP mean? We've seen that this is equivalent to AQFPAQFP being cyclic. We do an angle chasing galore again. What's the most natural way to prove that AQFPAQFP is cyclic? We might show that the sum of two opposite angles is 180180^{\circ}. Let's do it. Let's show that FQP+FAP=180\angle FQP + \angle FAP = 180^{\circ}. What is FQP\angle FQP? Look at FQB\triangle FQB-- FQP=180FQB=QFB+FBQ,\angle FQP = 180^{\circ} - \angle FQB = \angle QFB + \angle FBQ, where we've used the fact that QFB+FBQ+FQB=180.\angle QFB + \angle FBQ + \angle FQB = 180^{\circ}. Hence, we are reduced to proving that QFB+FBQ+FAP=180.\angle QFB + \angle FBQ + \angle FAP = 180^{\circ}. What is QFB\angle QFB? Notice that it is the same as DFB\angle DFB. We know that DFB=ACB\angle DFB = \angle ACB (why? hint: look at how we've proven AFE=ACB\angle AFE = \angle ACB; do the same arguments work?). Thus, we need to prove ACP+QFB+FAP=180.\angle ACP + \angle QFB + \angle FAP= 180^{\circ}. Now, let's try to manipulate FQB\angle FQB. Notice that it is the same angle as AQP,\angle AQP, which, by angles in the same arc, is equal to PCA\angle PCA. Also, notice that FAP\angle FAP is the same as BAP.\angle BAP. Hence, we need to prove that BCA+PCA+BAP=180.\angle BCA + \angle PCA + \angle BAP = 180^{\circ}. But, notice that BCA+PCA=BCP,\angle BCA + \angle PCA = \angle BCP, so we need to prove that BCP+BAP=180,\angle BCP + \angle BAP = 180^{\circ}, which is true (why?). Hence, we're done. \blacksquare

Now let's look at a very interesting property of cyclic quadrilaterals: the Simson line theorem.

Simson line theorem: Let DD be any point on the circumcircle of a triangle ABCABC. Let P,Q,RP, Q, R be the feet of perpendiculars from DD on BC,CA,ABBC, CA, AB respectively. Then, R,P,QR, P, Q are collinear. Conversely, if R,P,QR, P, Q are collinear, DD lies on the circumcircle of ABC\triangle ABC.

image link - http://s30.postimg.org/voesev3ap/Untitled.png image link - http://s30.postimg.org/voesev3ap/Untitled.png

Let's do the if part- if DD lies on the circumcircle of ABC,\triangle ABC, P,Q,RP, Q, R are collinear. There are several configurations possible here; WLOG we assume the one shown in the diagram. To show that R,P,QR, P, Q are collinear, it suffices to show that RPB=QPC\angle RPB = \angle QPC (why?). To do this angle chasing, we list all the cyclic quadrilaterals we have. By our assumption, ABDCABDC is a cyclic quadrilateral. Also, since DRRBDR \perp RB and DPBP,DP \perp BP, BRDPBRDP is cyclic with diameter BDBD (why?). Similarly, DRQCDRQC is also cyclic with diameter DRDR. Now, using these cyclic quadrilaterals, we have BPR=BDR\angle BPR = \angle BDR. From right angled BDR,\triangle BDR, we have BDR=90BRD\angle BDR = 90^{\circ} - \angle BRD. Similarly, QRC=90QCD\angle QRC = 90^{\circ} - \angle QCD. Hence, it suffices to show that DBR=DCQ.\angle DBR = \angle DCQ. Indeed, notice that DBR=180ABD=ACD=QCD,\angle DBR = 180^{\circ} - \angle ABD = \angle ACD = \angle QCD, so we're done. \blacksquare

The only if portion gets handled by the same arguments and is left to the reader.

Now, some exercises.

  • (IMO 2004 Problem 1): Let ABCABC be an acute-angled triangle with ABACAB\neq AC. The circle with diameter BCBC intersects the sides ABAB and ACAC at MM and NN respectively. Denote by OO the midpoint of the side BCBC. The bisectors of the angles BAC\angle BAC and MON\angle MON intersect at RR. Prove that the circumcircles of the triangles BMRBMR and CNRCNR have a common point lying on the side BCBC.

  • (IMO Shortlist 1997): Let ABCABC be a triangle. DD is a point on the side BCBC. The line ADAD meets the circumcircle again at XX. PP is the foot of the perpendicular from XX to ABAB, and QQ is the foot of the perpendicular from XX to ACAC. Show that the line PQPQ is a tangent to the circle with diameter XD XD if and only if AB=AC AB=AC.

  • (All Russian Olympiad 2007 Grade 9/10): The internal angle bisector of ABC\angle ABC in acute triangle ABCABC meets ACAC at B1B_1. The perpendicular from B1B_1 on BCBC meets the circumcircle of ABC\triangle ABC at KK (where KK is on minor arc AC^\widehat{AC}). The perpendicular from BB to AKAK meets ACAC at LL. Line BB1BB_1 meets the circumcircle of ABC\triangle ABC again at TT (TBT \neq B). Prove that K,L,TK, L, T are collinear.

  • (USA TSTST 2012 Problem 2): Let ABCDABCD be a quadrilateral with AC=BDAC = BD. Diagonals ACAC and BDBD meet at PP. Let ω1\omega_1 and O1O_1 denote the circumcircle and the circumcenter of triangle ABPABP. Let ω2\omega_2 and O2O_2 denote the circumcircle and circumcenter of triangle CDPCDP. Segment BCBC meets ω1\omega_1 and ω2\omega_2 again at SS and TT (other than BB and CC), respectively. Let MM and NN be the midpoints of minor arcs SP^\widehat {SP} (not including BB) and TP^\widehat {TP} (not including CC). Prove that MNMN is parallel to O1O2O_1O_2. (Note: The midpoint of arc SP^\widehat{SP} is the point which divides that arc into two equal parts. You could also consider it as the point where the internal angle bisector of SAP\angle SAP intersects ω1\omega_1 apart from AA. Similarly, consider NN as the point where the internal angle bisector of TCP\angle TCP meets ω2\omega_2 apart from CC. )

Note by Sreejato Bhattacharya
4 years, 12 months ago

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@@Calvin Lin Sir I'm not sure if this got added in the cyclic quadrilaterals skill judging by its url. Could you please verify?

Sreejato Bhattacharya - 4 years, 12 months ago

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Hi Sreejato. I see that this note has been added, and appears in the Cyclic Quadrilaterals Wiki page (listed at the bottom).

Could you copy and paste the first half (up to "Now, let's try a slightly harder problem."), into the "Write a summary" portion of the wiki page? That will allow others easy access to reading the points that you have presented. Thanks!

Calvin Lin Staff - 4 years, 12 months ago

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