Hi. This is an introductory note on cyclic quadrilaterals. I shall be writing a few more notes on basic geometry. I'm eager about other's feedbacks. Please tell me if this needs to be improved. I'm open to all suggestions. This note assumes basic knowledge of lines and triangles, typically found in a middle school syllabus.
Basically, a cyclic quadrilateral is a quadrilateral inscribed in a circle. In other words, quadrilateral 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 lying on a circle centered at .
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 lying on a circle with center . Let be any other point on the circle as shown in the figure below. Then, the angle subtended by at the center is defined as and the angle subtended by at the circumference is . We're going to prove that
Okay, so let's extend to any arbitrary point .
What can we say about ? It's isoceles! To be a bit more detailed, since and lie on a circle centered at so we must have From isoceles this implies Now, we use the fact that the angles of a triangle sum up to Now look at straight line We have (why?), so We apply similar arguments to to deduce that Now, notice that is equal to sum of angles and is equal to the sum of angles and Isn't the result obvious now? Indeed, 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 always equal to the sum of and ? The answer is no. Look at the following picture.
As you can see, is actually equal to the difference between and If we let slide to the other side of becomes the difference of and . Fortunately, these cases are handled similarly. We also have that 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.
The proof almost immediately follows from what we have proven just before. Suppose is the center of the circle. What does being a diameter of a circle mean? Of course, it mens its midpoint is the center of the circle. Then, since are collinear. We've just proven that Is the conclusion immediate now?
P.S: There is another nice proof of this fact without using any prior knowledge. Hint: reflect over to get a point Can you prove is a rectangle?
P.P.S: The converse of this statement holds too. If is any point such that lies on the circle with diameter . Can you prove this? If no, look below. The proof to the converse of the property below does the job for this one too.
This proof is also immediate. Let be the center of the circle. We have already proven that and . Combining them, the result immediately follows.
Okay, so this is an important property of cyclic quadrilaterals. If is a cyclic quadrilateral with its points lying on that order, (cyclic variations hold). Does the converse also hold true? If quadrilateral with its points lying in that order satisfies must it be cyclic?
Yes, it must be cyclic. Let's see why. Suppose it weren't cyclic. Then, the circle passing through (i.e. the circumcircle of ) would intersect line at points and where We know that Hence, Can we conclude from this? Indeed, we can. For any angle , two fixed points and a line passing through , there exists exactly one point on such that . Try to visualize this as a point moving on the line. As the point approaches increases, and as it moves away, the angle decreases. Hence, must be the same point as .
This proof is also pretty simple. Let's use what we proved earlier. Notice that and since they are angles in the same arc. Hence, What is the sum of the angles ? Well, the sum of the angles of a triangle is , so 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 is the midpoint and are the feet of perpendiculars from on respectively. Prove that
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 we already know the values of and so we can easily compute and show that However, let's not do it the hard way. Let's see what we're given to prove.
We wish to prove There are several approaches which look promising. We can angle chase and try to prove that However, apparently there's no "easy" path that seems to lead to the values of and Instead, let's prove that and lie on a circle centered at . Indeed, this is true: since lies on the circle with diameter . Since also lies on the circle with diameter . What is the center of the circle with diameter ? Of course it's ! So, we're done. Notice that in fact
Now, let's look at another trivial problem.
Problem: Prove that in the internal angle bisector of and perpendicular bisector of meet on the circumcircle of .
Basically, the internal angle bisector of is a line which divides into two equal parts. In other words, if is any point lying on the angle bisector of . The perpendicular bisector of of course, is a line perpendicular to passing through the midpoint of . In fact, if is any point lying on the perpendicular bisector of it is well known that (why? hint: ).
Now let's see what we're given to prove. Suppose the internal angle bisector of meets the perpendicular bisector of at . We are to prove lies on the circumcircle of or that quadrilateral is cyclic. Okay, this looks messy. Let's try something else.
If we can show that the point where the internal angle bisector of intersects the circumcircle again (apart from ) also lies on the perpendicular bisector of , we will be done. This idea seems a lot more feasible.
So, suppose the internal angle bisector of meets the circumcircle of again at . We need to prove that or equivalently, . We are given that and we need to prove that But this is obvious - being angles in the same arc, and So, we're done.
Let's now look at another trivial problem.
Problem: Let be the orthocenter of and the diametrically opposite point of on the circumcircle of . Prove that passes through the midpoint of .
Basically, is the point of intersection of the altitudes from the points to the opposite sides. In other words, is the unique point such that . Also, is the unique point such that is a diameter of the circumcircle of . What do we get from this? Well, from what we've seen before, and . Can we combine them somehow? Lines and are both perpendicular to . What does this mean? Well, they are parallel! So, . Similarly, we can show that . What can we say about quadrilateral ? 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 cuts diagonal at its midpoint, i.e. bisects . However, this is well known - the diagonals of a parallelogram bisect each other (why? hint: if intersects at ), so passes through the midpoint of . Notice that as a consequence, the midpoint of is in fact the midpoint of
Now, let's try a slightly harder problem.
Problem (IMO Shortlist 2010 G1): Let be an acute triangle with the feet of the altitudes lying on respectively. One of the intersection points of the line and the circumcircle is The lines and meet at point Prove that
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 is isoceles with . Let's try angle chasing. To show it shall be enough to show . But what is ? Well, let's look at the picture: is equal to which is equal to by angles in the same arc. Hence, we need to prove that . Let's see how many other angles are equal to . Look back at the first example- lie on a circle with diameter so is cyclic. What can we say about about this? From the fact that the opposite angles of a cyclic quadrilateral sum up to Now, what does this force to be equal to? Well, notice that This forces . Now, we need to prove that . Notice that . Now, what does mean? We've seen that this is equivalent to being cyclic. We do an angle chasing galore again. What's the most natural way to prove that is cyclic? We might show that the sum of two opposite angles is . Let's do it. Let's show that . What is ? Look at -- where we've used the fact that Hence, we are reduced to proving that What is ? Notice that it is the same as . We know that (why? hint: look at how we've proven ; do the same arguments work?). Thus, we need to prove Now, let's try to manipulate . Notice that it is the same angle as which, by angles in the same arc, is equal to . Also, notice that is the same as Hence, we need to prove that But, notice that so we need to prove that which is true (why?). Hence, we're done.
Now let's look at a very interesting property of cyclic quadrilaterals: the Simson line theorem.
Simson line theorem: Let be any point on the circumcircle of a triangle . Let be the feet of perpendiculars from on respectively. Then, are collinear. Conversely, if are collinear, lies on the circumcircle of .
Let's do the if part- if lies on the circumcircle of are collinear. There are several configurations possible here; WLOG we assume the one shown in the diagram. To show that are collinear, it suffices to show that (why?). To do this angle chasing, we list all the cyclic quadrilaterals we have. By our assumption, is a cyclic quadrilateral. Also, since and is cyclic with diameter (why?). Similarly, is also cyclic with diameter . Now, using these cyclic quadrilaterals, we have . From right angled we have . Similarly, . Hence, it suffices to show that Indeed, notice that so we're done.
The only if portion gets handled by the same arguments and is left to the reader.
Now, some exercises.
(IMO 2004 Problem 1): Let be an acute-angled triangle with . The circle with diameter intersects the sides and at and respectively. Denote by the midpoint of the side . The bisectors of the angles and intersect at . Prove that the circumcircles of the triangles and have a common point lying on the side .
(IMO Shortlist 1997): Let be a triangle. is a point on the side . The line meets the circumcircle again at . is the foot of the perpendicular from to , and is the foot of the perpendicular from to . Show that the line is a tangent to the circle with diameter if and only if .
(All Russian Olympiad 2007 Grade 9/10): The internal angle bisector of in acute triangle meets at . The perpendicular from on meets the circumcircle of at (where is on minor arc ). The perpendicular from to meets at . Line meets the circumcircle of again at (). Prove that are collinear.
(USA TSTST 2012 Problem 2): Let be a quadrilateral with . Diagonals and meet at . Let and denote the circumcircle and the circumcenter of triangle . Let and denote the circumcircle and circumcenter of triangle . Segment meets and again at and (other than and ), respectively. Let and be the midpoints of minor arcs (not including ) and (not including ). Prove that is parallel to . (Note: The midpoint of arc 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 intersects apart from . Similarly, consider as the point where the internal angle bisector of meets apart from . )
(More to come soon.) Feel free to discuss these problems in the comments.
Okay, so far for now. I'm exhausted now; will complete the rest of this note tomorrow. Please wait, and provide feedback for the note so far. :)
I'll include more example and practice problems from past olympiads, in case anyone's interested.