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Basically, a cyclic quadrilateral is a quadrilateral inscribed in a circle. In other words, quadrilateral $$ABCD$$ 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 $$A, B, C, D$$ lying on a circle centered at $$O$$.

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 $$A, B$$ lying on a circle with center $$O$$. Let $$C$$ be any other point on the circle as shown in the figure below. Then, the angle subtended by $$\widehat{AB}$$ at the center is defined as $$\angle AOB,$$ and the angle subtended by $$\widehat{AB}$$ at the circumference is $$\angle ACB$$. We're going to prove that $$\angle AOB = 2 \angle ACB.$$

Okay, so let's extend $$CO$$ to any arbitrary point $$X$$.

What can we say about $$\triangle OAC$$? It's isoceles! To be a bit more detailed, since $$A$$ and $$C$$ lie on a circle centered at $$O,$$ so we must have $$OA=OC.$$ From isoceles $$\triangle OAC,$$ this implies $$\angle OAC= \angle OCA.$$ Now, we use the fact that the angles of a triangle sum up to $$180^{\circ}.$$ $\angle OAC + \angle OCA + \angle AOC = 180^{\circ} \implies \angle AOC = 180^{\circ} - 2\angle OCA.$ Now look at straight line $$COX.$$ We have $$\angle COA + \angle AOX = 180^{\circ}$$ (why?), so $\angle AOX = 180^{\circ} - \angle AOC = 2 \angle OCA.$ We apply similar arguments to $$\triangle OAB$$ to deduce that $$\angle XOB = 2\angle OCB.$$ Now, notice that $$\angle AOB$$ is equal to sum of angles $$\angle AOX + \angle XOB,$$ and $$\angle ACB$$ is equal to the sum of angles $$\angle ACO$$ and $$\angle BCO.$$ Isn't the result obvious now? Indeed, $\angle AOB = \angle AOX + \angle XOB = 2 ( \angle ACO + \angle OCB) = 2 \angle ACB.$ 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 $$\angle AOB$$ always equal to the sum of $$\angle AOX$$ and $$\angle XOB$$? The answer is no. Look at the following picture.

As you can see, $$\angle AOB$$ is actually equal to the difference between $$\angle BOX$$ and $$\angle AOX.$$ If we let $$C$$ slide to the other side of $$O,$$ $$\angle AOB$$ becomes the difference of $$\angle AOX$$ and $$\angle BOX$$. Fortunately, these cases are handled similarly. We also have that $$\angle ACB = \angle OCB - \angle OCA.$$ 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 $$AB$$ is a diameter of a circle and $$C$$ any other point on that circle, $$\angle ACB = 90^{\circ}$$.

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

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

P.P.S: The converse of this statement holds too. If $$C$$ is any point such that $$\angle ACB = 90^{\circ},$$ $$C$$ lies on the circle with diameter $$AB$$. 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 $$A, B$$ on a circle and any other point $$C$$ on that circle not lying between $$A$$ and $$B$$. As $$C$$ varies along the circumference of the circle, $$\angle ACB$$ remains constant. Alternatively, if $$C$$ and $$D$$ are two points on that circle, neither of them lying between $$A$$ and $$B,$$ $$\angle ACB = \angle ADB.$$

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

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

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

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

This proof is also pretty simple. Let's use what we proved earlier. Notice that $$\angle CAB = \angle CDB$$ and $$\angle ACB = \angle ADB$$ since they are angles in the same arc. Hence, $$\angle BDC + \angle ABC = \angle BDC + \angle ADB + \angle ABC = \angle ABC + \angle BAC + \angle BCA.$$ What is the sum of the angles $$\angle CAB + \angle ABC + \angle BCA$$? Well, the sum of the angles of a triangle is $$180^{\circ}$$, so $$\angle CAB + \angle ABC + \angle BCA = 180^{\circ}.$$ 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 $$\triangle ABC,$$ $$M$$ is the midpoint and $$E, F$$ are the feet of perpendiculars from $$B, C$$ on $$AC, AB$$ respectively. Prove that $$ME = MF.$$

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 $$\triangle ABC,$$ we already know the values of $$CM, CE$$ and $$\cos C,$$ so we can easily compute $$ME$$ and show that $$ME=MC.$$ However, let's not do it the hard way. Let's see what we're given to prove.

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

Now, let's look at another trivial problem.

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

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

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

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

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

Let's now look at another trivial problem.

Problem: Let $$H$$ be the orthocenter of $$\triangle ABC$$ and $$A'$$ the diametrically opposite point of $$A$$ on the circumcircle of $$\triangle ABC$$. Prove that $$HA'$$ passes through the midpoint of $$BC$$.

Basically, $$H$$ is the point of intersection of the altitudes from the points to the opposite sides. In other words, $$H$$ is the unique point such that $$BH \perp AC, AH \perp BC, CH \perp AB$$. Also, $$A'$$ is the unique point such that $$AA'$$ is a diameter of the circumcircle of $$\triangle ABC$$. What do we get from this? Well, from what we've seen before, $$A'B \perp AB$$ and $$A'C \perp AC$$. Can we combine them somehow? Lines $$A'B$$ and $$CH$$ are both perpendicular to $$AB$$. What does this mean? Well, they are parallel! So, $$A'B \parallel CH$$. Similarly, we can show that $$A'C \parallel BH$$. What can we say about quadrilateral $$BA'CH$$? 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 $$HA'$$ cuts diagonal $$BC$$ at its midpoint, i.e. bisects $$BC$$. However, this is well known - the diagonals of a parallelogram bisect each other (why? hint: if $$HA'$$ intersects $$BC$$ at $$M,$$ $$\triangle BHM \cong \triangle A'MC$$), so $$HA'$$ passes through the midpoint of $$BC$$. Notice that as a consequence, the midpoint of $$BC$$ is in fact the midpoint of $$HA'.$$

Now, let's try a slightly harder problem.

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

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

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

Let's do the if part- if $$D$$ lies on the circumcircle of $$\triangle ABC,$$ $$P, Q, R$$ are collinear. There are several configurations possible here; WLOG we assume the one shown in the diagram. To show that $$R, P, Q$$ are collinear, it suffices to show that $$\angle RPB = \angle QPC$$ (why?). To do this angle chasing, we list all the cyclic quadrilaterals we have. By our assumption, $$ABDC$$ is a cyclic quadrilateral. Also, since $$DR \perp RB$$ and $$DP \perp BP,$$ $$BRDP$$ is cyclic with diameter $$BD$$ (why?). Similarly, $$DRQC$$ is also cyclic with diameter $$DR$$. Now, using these cyclic quadrilaterals, we have $$\angle BPR = \angle BDR$$. From right angled $$\triangle BDR,$$ we have $$\angle BDR = 90^{\circ} - \angle BRD$$. Similarly, $$\angle QRC = 90^{\circ} - \angle QCD$$. Hence, it suffices to show that $$\angle DBR = \angle DCQ.$$ Indeed, notice that $$\angle DBR = 180^{\circ} - \angle ABD = \angle ACD = \angle QCD,$$ 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 $$ABC$$ be an acute-angled triangle with $$AB\neq AC$$. The circle with diameter $$BC$$ intersects the sides $$AB$$ and $$AC$$ at $$M$$ and $$N$$ respectively. Denote by $$O$$ the midpoint of the side $$BC$$. The bisectors of the angles $$\angle BAC$$ and $$\angle MON$$ intersect at $$R$$. Prove that the circumcircles of the triangles $$BMR$$ and $$CNR$$ have a common point lying on the side $$BC$$.

• (IMO Shortlist 1997): Let $$ABC$$ be a triangle. $$D$$ is a point on the side $$BC$$. The line $$AD$$ meets the circumcircle again at $$X$$. $$P$$ is the foot of the perpendicular from $$X$$ to $$AB$$, and $$Q$$ is the foot of the perpendicular from $$X$$ to $$AC$$. Show that the line $$PQ$$ is a tangent to the circle with diameter $$XD$$ if and only if $$AB=AC$$.

• (All Russian Olympiad 2007 Grade 9/10): The internal angle bisector of $$\angle ABC$$ in acute triangle $$ABC$$ meets $$AC$$ at $$B_1$$. The perpendicular from $$B_1$$ on $$BC$$ meets the circumcircle of $$\triangle ABC$$ at $$K$$ (where $$K$$ is on minor arc $$\widehat{AC}$$). The perpendicular from $$B$$ to $$AK$$ meets $$AC$$ at $$L$$. Line $$BB_1$$ meets the circumcircle of $$\triangle ABC$$ again at $$T$$ ($$T \neq B$$). Prove that $$K, L, T$$ are collinear.

• (USA TSTST 2012 Problem 2): Let $$ABCD$$ be a quadrilateral with $$AC = BD$$. Diagonals $$AC$$ and $$BD$$ meet at $$P$$. Let $$\omega_1$$ and $$O_1$$ denote the circumcircle and the circumcenter of triangle $$ABP$$. Let $$\omega_2$$ and $$O_2$$ denote the circumcircle and circumcenter of triangle $$CDP$$. Segment $$BC$$ meets $$\omega_1$$ and $$\omega_2$$ again at $$S$$ and $$T$$ (other than $$B$$ and $$C$$), respectively. Let $$M$$ and $$N$$ be the midpoints of minor arcs $$\widehat {SP}$$ (not including $$B$$) and $$\widehat {TP}$$ (not including $$C$$). Prove that $$MN$$ is parallel to $$O_1O_2$$. (Note: The midpoint of arc $$\widehat{SP}$$ 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 $$\angle SAP$$ intersects $$\omega_1$$ apart from $$A$$. Similarly, consider $$N$$ as the point where the internal angle bisector of $$\angle TCP$$ meets $$\omega_2$$ apart from $$C$$. )

Note by Sreejato Bhattacharya
2 years, 5 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? · 2 years, 5 months ago

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! Staff · 2 years, 5 months ago