Theorems Related to Cyclic Quadrilateral

In plane geometry, any three non-collinear points $A, B, C$ determine a unique circle passing through them (their circumcircle, whose center is the intersection of the perpendicular bisectors of the sides). However, if a fourth point $D$ is chosen at random in the plane, it will generally not lie on this circumcircle.

• Concyclic Points (সমবৃত্তস্থ বিন্দু): Points that all lie on the circumference of the same circle are called concyclic points.
• Cyclic Quadrilateral (বৃত্তস্থ চতুর্ভুজ): A quadrilateral whose all four vertices lie on the circumference of a circle is called a cyclic quadrilateral. The circle is called the circumcircle of the quadrilateral, and the quadrilateral is said to be inscribed in the circle.

Let $ABCD$ be a cyclic quadrilateral inscribed in a circle with center $O$:

Follow the 5-step classical Euclidean proof required in Madhyamik examinations:

1. Given (প্রদত্ত): Let $ABCD$ be a cyclic quadrilateral inscribed in a circle with center $O$.
2. To Prove (প্রমাণ করতে হবে যে):
   (i) $\angle ABC + \angle ADC = 2\text{ right angles} = 180^\circ$
   (ii) $\angle DAB + \angle BCD = 2\text{ right angles} = 180^\circ$
3. Construction (অঙ্কন): Join radii $OA$ and $OC$.
4. Proof (প্রমাণ):
   Consider the arc $ADC$ (the arc not containing vertex $B$):
   This arc subtends the central angle $\angle AOC$ at the center $O$, and subtends the inscribed angle $\angle ABC$ at the circumference.
   By Theorem 34 (central angle is double the inscribed angle): $$\text{reflex } \angle AOC = 2\angle ABC \quad \dots \text{(Equation 1)}$$
   Now consider the arc $ABC$ (the arc not containing vertex $D$):
   This arc subtends the central angle $\angle AOC$ at the center $O$, and subtends the inscribed angle $\angle ADC$ at the circumference.
   By Theorem 34: $$\angle AOC = 2\angle ADC \quad \dots \text{(Equation 2)}$$
   Adding Equation 1 and Equation 2: $$\angle AOC + \text{reflex } \angle AOC = 2\angle ADC + 2\angle ABC$$ $$\angle AOC + \text{reflex } \angle AOC = 2(\angle ADC + \angle ABC)$$
   Notice that the angle around the point $O$ forms a complete circle ($4\text{ right angles} = 360^\circ$): $$\angle AOC + \text{reflex } \angle AOC = 360^\circ$$
   Therefore: $$2(\angle ABC + \angle ADC) = 360^\circ$$ $$\mathbf{\angle ABC + \angle ADC = 180^\circ} \quad (2\text{ right angles})$$
   Since the sum of all four interior angles of any convex quadrilateral is $360^\circ$: $$\angle DAB + \angle ABC + \angle BCD + \angle ADC = 360^\circ$$ $$(\angle DAB + \angle BCD) + (\angle ABC + \angle ADC) = 360^\circ$$ $$(\angle DAB + \angle BCD) + 180^\circ = 360^\circ$$ $$\mathbf{\angle DAB + \angle BCD = 180^\circ}$$
5. Conclusion: The opposite angles of cyclic quadrilateral $ABCD$ are supplementary. $\quad \blacksquare$ (Q.E.D.)

Proof by Contradiction (Reductio ad Absurdum): Suppose $ABCD$ is a quadrilateral in which $\angle B + \angle D = 180^\circ$, but vertices $A, B, C, D$ are not concyclic. Draw the unique circle passing through the three non-collinear points $A, B, C$. If $D$ does not lie on this circle, the circle must intersect line $AD$ (or $AD$ produced) at some other point $E$. Join $CE$. Then quadrilateral $ABCE$ is cyclic $\implies \angle B + \angle AEC = 180^\circ$. Comparing with $\angle B + \angle D = 180^\circ$, we get $\angle AEC = \angle D = \angle ADC$. In $\triangle CDE$, this implies that an exterior angle equals an interior opposite angle, which is impossible unless $E$ coincides with $D$. Thus, $D$ must lie on the circle, proving that $A, B, C, D$ are concyclic.

1. Let $ABCD$ be a cyclic quadrilateral. Produce side $AB$ beyond $B$ to a point $E$.
2. Then $\angle CBE$ is the exterior angle at vertex $B$.
3. Since $ABE$ is a straight line segment, angles $\angle ABC$ and $\angle CBE$ form a linear pair: $$\angle ABC + \angle CBE = 180^\circ \quad \dots \text{(Equation 1)}$$
4. By Theorem 38, the opposite angles of cyclic quadrilateral $ABCD$ are supplementary: $$\angle ABC + \angle ADC = 180^\circ \quad \dots \text{(Equation 2)}$$
5. Equating Equation 1 and Equation 2: $$\angle ABC + \angle CBE = \angle ABC + \angle ADC$$
6. Subtract $\angle ABC$ from both sides: $$\mathbf{\angle CBE = \angle ADC}$$

Thus, the exterior angle at vertex $B$ is identically equal to the interior opposite angle at vertex $D$. This property applies to any side produced at any of the four vertices.

Theorem: Prove that any cyclic parallelogram is a rectangle.

1. Let $ABCD$ be a cyclic parallelogram.
2. Because $ABCD$ is a parallelogram, its opposite angles are equal: $$\angle A = \angle C \quad \text{and} \quad \angle B = \angle D \quad \dots \text{(Property of Parallelogram)}$$
3. Because $ABCD$ is cyclic, its opposite angles are supplementary (Theorem 38): $$\angle A + \angle C = 180^\circ$$
4. Substitute $\angle C = \angle A$: $$\angle A + \angle A = 180^\circ \implies 2\angle A = 180^\circ \implies \mathbf{\angle A = 90^\circ}$$
5. Since one interior angle of parallelogram $ABCD$ is $90^\circ$, all four angles are $90^\circ$: $$\angle A = \angle B = \angle C = \angle D = 90^\circ$$
6. A parallelogram with a right angle is, by definition, a rectangle. $\quad \blacksquare$

Theorem: Prove that a cyclic trapezium is isosceles (its non-parallel sides are equal, and base angles are equal).

1. Let $ABCD$ be a cyclic trapezium in which $AB \parallel DC$, and $AD, BC$ are non-parallel sides.
2. Because $AB \parallel DC$ and transversal $AD$ intersects them, consecutive interior angles are supplementary: $$\angle A + \angle D = 180^\circ \quad \dots \text{(Equation 1)}$$
3. Because $ABCD$ is a cyclic quadrilateral, opposite angles are supplementary (Theorem 38): $$\angle B + \angle D = 180^\circ \quad \dots \text{(Equation 2)}$$
4. Comparing Equation 1 and Equation 2: $$\angle A + \angle D = \angle B + \angle D \implies \mathbf{\angle A = \angle B}$$
5. Similarly, $\angle C = \angle D$. Since the base angles are equal, the trapezium is symmetrical: the non-parallel sides are equal ($AD = BC$), and the diagonals are equal ($AC = BD$).
6. Therefore, a cyclic trapezium is strictly an isosceles trapezium. $\quad \blacksquare$

A rhombus is a parallelogram with all four sides equal. By Rider 1, a cyclic parallelogram is a rectangle (all angles $90^\circ$). A rectangle with all four sides equal is a square. Therefore, every cyclic rhombus is inevitably a square!

A classic 3-mark or 4-mark Madhyamik geometry problem involves two circles intersecting at two points $P$ and $Q$:

1. Draw the common chord $PQ$.
2. Through $P$ and $Q$, two straight line segments $APB$ and $CQD$ are drawn, cutting the first circle at $A, C$ and the second circle at $B, D$.
3. In the first circle, $APQC$ is a cyclic quadrilateral $\implies \angle PAC + \angle PQC = 180^\circ$.
4. Since $CQD$ is a straight line, $\angle PQC + \angle PQD = 180^\circ \implies \mathbf{\angle PQD = \angle PAC}$ (exterior angle of cyclic quadrilateral).
5. In the second circle, $PBQD$ is a cyclic quadrilateral $\implies \angle PBD + \angle PQD = 180^\circ$.
6. Substitute $\angle PQD = \angle PAC$: $$\angle PBD + \angle PAC = 180^\circ \implies \mathbf{\angle B + \angle A = 180^\circ}$$
7. Since consecutive interior angles between lines $AC$ and $BD$ cut by transversal $AB$ add up to $180^\circ$: $$\mathbf{AC \parallel BD}$$
8. This proves that the line segments joining the endpoints are strictly parallel!