কখন একটি চতুর্ভুজ বৃত্তস্থ হয় ?

কোন চতুর্ভুজের শীর্ষবিন্দু ৪টি দিয়ে একটি বৃত্ত আঁকা গেলে সেই চতুর্ভুজকে বৃত্তস্থ চতুর্ভুজ বলে। বৃত্তস্থ চতুর্ভুজ নিয়ে কথা বলার আগে ত্রিভুজ নিয়ে কিছু বলি।

তাহলে কী চতুর্ভুজের ক্ষেত্রেও এই বৈশিষ্ট্য বিদ্যমান?

না। সকল চতুর্ভুজ বৃত্তস্থ চতুর্ভুজ না। কেন না? (উফফ, এত প্রশ্ন!)

কেননা, একটি ত্রিভুজের পরিবৃত্ত অনন্য হওয়ার ফলে, বৃত্তস্থ চতুর্ভুজের যেকোন তিন বিন্দুর বৃত্ত অপর বিন্দুগামী হবে। সুতরাং, সব চতুর্ভুজ বৃত্তস্থ না। কিছু বিশেষ শর্তসাপেক্ষে জন্য একটি চতুর্ভুজ বৃত্তস্থ হবে।

একটি শর্ত হলো,বৃত্তস্থ চতুর্ভুজের বিপরীত কোণদ্বয়ের সমষ্টি $180^\circ$।

আমরা জানি, কেন্দ্রস্থ কোণ বৃত্তস্থ কোণের দ্বিগুণ হয়। এটি ব্যবহার করে আমরা বৃত্তস্থ চতুর্ভুজের এই বৈশিষ্ট্য প্রমাণ করব। চিত্র অনুযায়ী,
$$\begin{align}\angle ABP &=\frac{1}{2}\angle AOP\left(1\right)........\left(i\right)\\ \angle AQP&=\frac{1}{2}\angle AOP\left(2\right)........\left(ii\right)\\ [\left(i\right)+\left(ii\right)]\angle ABP+\angle AQP &=\frac{1}{2}\left(\angle AOP(1)+\angle AOP(2)\right) \end{align}$$ কিন্তু, $\angle AOP\left(1\right)+\angle AOP\left(2\right)=360^\circ$.

অতএব, $\angle ABP+\angle AQP=\frac{1}{2}\left(360^\circ\right)=180^\circ$

আবার, উপরের স্টেট্মেন্টের বিপরিতটাও সত্য। অর্থাৎ, কোন চতুর্ভুজের বিপরীত কোণদ্বয়ের সমষ্টি $180º$ হলে চতুর্ভুজটি বৃত্তস্থ হয়।

আমরা বিরোধ বা কন্ট্রাডিকশনের মাধ্যমে প্রমাণ করব।

ধরে নেই, $\angle ABC+\angle AD'C=180^\circ$। $D$ বিন্দু $CD'$ ও $\triangle ABC$ এর পরিবৃত্তের ছেদবিন্দু। উপরের প্রমাণ অনুযায়ী, যেহেতু $ABCD$ একটি বৃত্তস্থ চতুর্ভুজ, $\angle ABC+\angle ADC=180^\circ$। সুতরাং, $\angle ADC=\angle AD'C$। কিন্তু, $D$ ও $D'$ আলাদা বিন্দু হলে এটা সম্ভব না। কেননা বহিঃস্থ কোণ বিপরীত অন্তস্থ কোণদ্বয়ের সমষ্টির সমান। টা ডা! আমাদের প্রমাণ এখন সম্পূর্ণ।

গুড লাক এবং জিও অন।

Ancient mathematicians were fascinated with geometric coincidences. Let's get a feel for this ourselves.

I drew a triangle in the sand, and noticed that it was possible to draw a perfect circle through the the 3 points. Wow!

And, it always looks like some rotation of this:

It would be nice to understand more general quadrilaterals, like this one:

We say that a quadrilateral is cyclic if it's possible to draw a circle through its four points. Some quadrilaterals are obviously not cyclic, like this one:

It turns out that there's a beautiful theorem which characterizes cyclic quadrilaterals.

Theorem: A quadrilateral is cyclic if and only if opposite pairs of angles add up to 180 degrees.

Proof:

Let the center of the circle be $O$. Then draw $OA$ and $OC$. By the Inscribed Angle Theorem, $m\angle ABC = \frac{1}{2} m\angle AOC$ and $m\angle ADC = \frac{1}{2} m\angle COA$ (where $m\angle COA = 360º -\angle AOC$). Since $m\angle COA + m\angle AOC = 360º$, $m\angle ABC + m\angle ADC = \frac{360º}{2} = 180º$. Our theorem has been proven!

We can use the part that we just proved to finish the proof of the theorem. Let's do so.

Suppose there exists a quadrilateral with opposite angles that sum up to $180º$, but is not cyclic. [This technique is called Proof by Contradiction – assuming that what we wish to prove is false and then finding an issue that contradicts our assumption] That quadrilateral is $ABCD'$ below. Note that $ABCD$ is cyclic, and $D'$ is on $CD$.

Since it is not cyclic, the circumcircle of $\triangle ABC$ must not also go through $D'$. Now locate point $D$ on $CD'$. By our assumption, $m \angle AD'C + m \angle ABC = 180º$. However, our proof of the first part demonstrates that $\angle ADC$ is also supplementary to $\angle ABC$ and therefore $\angle ABC \cong \angle AD'C$.

The exterior angle of $\triangle ADD'$ cannot be the same size as an interior angle. This contradiction proves the theorem. Thus a quadrilateral with opposite angles summing up to $180º$ means that it is cyclic.

WOOT! That was a lot. Also, the waiting period is now over. After 8 months (?) you now have the proof of both the "if" and the "only if" parts of the theorem :D

There are several fundamental criteria that help determine if a quadrilateral is cyclic.

First, a quadrilateral is cyclic iff opposite angles sum to $180^\circ$. This is because an inscribed angle of a circle equals half the measure of the arc, so $$\angle ABC+\angle ADC=\dfrac{\widehat{ADC}}2+\dfrac{\widehat{ABC}}2=\dfrac{360^\circ}2=180^\circ.$$

A quadrilateral $ABCD$ is also cyclic iff $\angle ABD=\angle ACD$. This equality holds in cyclic quads because both angles subtend the same arc $\widehat{AD}$. By examining each of the other three arcs, similar equality conditions can be obtained.

[The "if" directions were proven earlier here. The "only if" direction can be proven many ways; for instance, we may use the "angle-arc inequalities" discussed earlier.]

These are the two most important criteria to remember when starting off with cyclic quadrilaterals. Well, why are these important, you may ask? Well, cyclic quadrilaterals allow us to shift between different sets of angle equalities that would be very difficult to work out otherwise.

Try it out on the following simple application.

In the diagram $\triangle ABC$ be a right triangle with $\angle B=90^\circ$. Point $D$ is an arbitrary point on leg $\overline{BC}$, and $E$ is the foot of the perpendicular from $D$ to $\overline{AC}$. We wish to prove that $\angle DBE=\angle DAE$.

Using the cyclic quad makes the problem easy and borderline trivial. Imagine trying to prove the angle equality otherwise!

In Olympiad Geometry, cyclic quadrilaterals (and spotting them) are crucial. To give you a glimpse as to how important cyclic quads are, here are some problems that appeared on actual olympiads that employ them in some way or another.

(USAMO 1993) Let $\, ABCD \,$ be a convex quadrilateral such that diagonals $\, AC \,$ and $\, BD \,$ intersect at right angles, and let $\, E \,$ be their intersection. Prove that the reflections of $\, E \,$ across $\, AB, \, BC, \, CD, \, DA \,$ are concyclic.

(Sharygin 2012) On side $AC$ of triangle $ABC$ an arbitrary point is selected $D$. The tangent in $D$ to the circumcircle of triangle $BDC$ meets $AB$ in point $C_{1}$; point $A_{1}$ is defined similarly. Prove that $A_{1}C_{1}\parallel AC$.

(Bulgaria 1993) A parallelogram $ABCD$ with an acute angle $BAD$ is given. The bisector of $\angle BAD$ intersects $CD$ at point $L$, and the line $BC$ at point $K$. Prove that the circumcenter of $\triangle LCK$ lies on the circumcircle of $\triangle BCD$.

(Sharygin 2009) Let $ABC$ be a triangle. Points $M$, $N$ are the projections of $B$ and $C$ to the bisectors of angles $C$ and $B$ respectively. Prove that line $MN$ intersects sides $AC$ and $AB$ in their points of contact with the incircle of $ABC$.

(USAMO 2008) Let $ABC$ be an acute, scalene triangle, and let $M$, $N$, and $P$ be the midpoints of $\overline{BC}$, $\overline{CA}$, and $\overline{AB}$, respectively. Let the perpendicular bisectors of $\overline{AB}$ and $\overline{AC}$ intersect ray $AM$ in points $D$ and $E$ respectively, and let lines $BD$ and $CE$ intersect in point $F$, inside of triangle $ABC$. Prove that points $A$, $N$, $F$, and $P$ all lie on one circle.

In short, learn how to spot cyclic quads and use them effectively and a whole new world of geometry problems will open at your fingertips.

When is a quadrilateral cyclic?

In general, a polygon is cyclic if there exists a circle passing through all the vertices of the polygon.

So, before we look at quadrilateral, let's start with small cases, i.e. triangles.

Why? Note that every three noncollinear points determine a unique circle. Why? We'll show this result.

Suppose firstly you have two points, $A$ and $B$.

Now, consider a third point, $C$, that is not collinear with $A$ and $B$. The locus of point equidistant to $A$ and $B$, and the locus of point equidistant to $A$ and $C$, clearly intersect at one point, $O$. This point will be equidistant to $A$, $B$ and $C$ and thus the circle we want is the circle with center $O$ that passes through $A$. We also proved that the circle is unique.

So, every triangle with cyclic. What if we add in another point, say $D$.

Why? Since the circle passing through $A, B, C, D$ must pass through $A, B$ and $C$. Thus, $D$ must lie on the circumcircle of $ABC$.

So, how do I determine whether a quadrilateral is cyclic?

This post gave a proof to the facts that

1. The opposite angles of a cyclic quadrilateral adds up to $180^{\circ}$.

2. In the diagram above, $\angle CDB = \angle CAB$, $\angle BDA = \angle ACB$, $\angle DCA = \angle DBA$, $\angle DAC = \angle DBC$.

Now, these conditions are if and only if, i.e. they hold conversely. In the second proposition, if any of the angle equalities hold, then the quadrilateral is cyclic.

We'll show that if the opposite angles of the quadrilateral adds up to $180^{\circ}$, then the quadrilateral is cyclic.

In the diagram above, assume that $\angle ABC + \angle CDA = 180^{\circ}$. Let $E$ be the second intersection of line $AD$ and the circumcircle of triangle $ABC$. We know that $\angle AEC = 180^{\circ} - \angle ABC = \angle ADC$, so $D = E$, and thus $ABCD$ is cyclic.

Now you try to prove the other proposition!

This proves the second proposition.

Thanks for reading!

Let $ABC$ be a triangle, and $D, E, F$ lie on $BC, AC, AB$ respectively. Let $\odot AEF$ and $\odot BDF$ intersect at $P$. We're going to show that in fact the third circle ($\odot CDE$) ALSO goes through $P$!

Here's a diagram for reference:

Ok, let's start now that we have a nice picture to look off. :)

We want to show that $P$ lies on $\odot CDE$.

The best way to say that $P$ lies on $\odot CDE$ is probably to say that $PCDE$ is cyclic.

Good! A quadrilateral is cyclic if and only if its opposite angles add to $180^\circ$. Now, let's try to show that this is true. To prove this, we need to use some facts about the cyclic quadrilaterals $AFPE$ and $BFPE$.

Once again, all we used was that opposite angles add to $180^\circ$.

Once again, all we used was the fact that opposite angles in a quadrilateral add to $180^\circ$. Now let's finish our proof. Let's list out the equations that we have now:

$\angle EAF + \angle FPE = 180^\circ.$ $\angle FBD + \angle FPD = 180^\circ.$

Let's find $\angle EPD$ now. Note that $\angle EPD = 360^\circ - \angle FPE - \angle FPD$. Let's solve for $\angle FPE$ and $\angle FPD$ using our above equations.

$\angle FPE = 180^\circ - \angle EAF.$ $\angle FPD = 180^\circ - \angle FBD.$

Let's plug these in to our equation for $\angle EPD.$ This gives us that

$\angle EPD = 360^\circ - \angle FPE - \angle FPD$

$= 360^\circ - (180^\circ - \angle EAF) - (180^\circ - \angle FBD)$

$= \angle EAF + \angle FBD.$

We already found $\angle EPD$ in terms of other angles.

$\angle EPD = \angle EAF + \angle FBD.$

Finally, all we need to show is that $\angle EAF + \angle FBD + \angle ECD = 180^\circ.$ But if you look closely, these angles are in fact the same as the three angles of $\triangle ABC$. So of course they add to $180^\circ$! Thus, we are done.

$P$ is often called the Miquel Point of the points $D, E, F$.

When does a Quadrilateral become Cyclic?(Incomplete)

Fundamental definition:

A quadrilateral $ABCD$ is $\textbf{cyclic}$ if there is a circle passing through four vertices of the quadrilateral.

Let's move on.

This post is not about basics of cyclic quads; it's about the fun we can have with them.There are a lot of $indicators$ (i.e. theorems, lemmas, etc.) which indicate that a particular quadrilateral is cyclic.

Now you won't see cyclic quadrilaterals tagged as $\textbf{Cyclic Quads}$ in your exams or wherever you use them in life, so we will try to find them by using the primary indicator, angle chasing.

However,

and angle chasing alone doesn't happen to solve it.

Let's take this video and add even more life. This person is very enthusiastic!

(There is still a gap here to contribute to!)

Quadricyclicity given side lengths?

Here's one you might have to think about for a while:

Also, so either way, let's assume there is such a construction. If we didn't know the answer, we would start like this and then just hope for the best.

So assume quadrilateral $ABCD$ is arranged like so and we label point $M$ lying on line $\overline{AD}$. such that $\angle DCM = \angle BAC$. Now, let's start making some observations.

First, it is noteworthy that $$\angle ABC = 180^\circ - \angle ADC = \angle CDM,$$ making $\triangle ABC \sim \triangle CDM$.

Due to this pair of similar triangles, we have $$\frac {DM}{b} = \frac {c}{a} \implies DM = \frac {bc}{a}.$$Also, $\frac {CM}{f} = \frac {c}{a}$.

This circle is also called an Apollonian circle.

Think some more ... and only click that button when you think you've gotten it.

The construction

• Draw $AD=d$ and extend it to $M$ so that $DM = \frac {bc}{a}$.

• $C$ is the intersection of the two circles: the one with center $D$ and radius $c$ and the Apollonian.

• $B$ is the intersection of the circle with center $A$ and radius $a$ and the circle with center $C$ and radius $b$.

Now, we should be done! The proof that this works is left as an exercise for the reader.