Art of Problem Solving Volume 2: and Beyond

Cyclic Quadrilaterals

4.1 Properties of Cyclic Quadrilaterals

• A cyclic quadrilateral is a quadrilateral that can be inscribed in a circle.
• Property 1: The sum of opposite angles in a cyclic quadrilateral is always 180°.
  • LA + LC = 180°
  • LB + LD = 180°
  • Proof: Angles A and C are inscribed angles.
    • LA + LC = \frac{BCD}{2} + \frac{BAD}{2} = \frac{BCD + BAD}{2} = \frac{360°}{2} = 180°
• Property 2: When diagonals of a cyclic quadrilateral are drawn, four pairs of equal angles are formed.
  • These angles are equal because they are inscribed angles subtending the same arc.
    • LABD = LACD
    • LBAC = LBDC
    • LACB = LADB
    • LCBD = LCAD

4.2 Finding Cyclic Quadrilaterals

• Method 1: If the sum of a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.
  • If one pair of opposite angles sums to 180°, the other pair also must, since the sum of all four angles in a quadrilateral is 360°.
  • This is the easiest and most common method to prove a quadrilateral is cyclic.
• Proof:
  • Given quadrilateral ABCD, assume LB + LD = 180°.
  • Draw the circumcircle of triangle ABC and show that the circle must pass through D.
  • Prove D cannot be inside or outside the circle.
• Case 1: D is inside the circle
  • Let EF = \theta, AEC = \alpha, and ABC = \beta.
  • Since AEC and ABC make up the whole circle, alpha + beta = 360°.
  • LB + LD = \frac{\beta}{2} + \frac{\alpha + \theta}{2} = \frac{\beta + \alpha}{2} + \frac{\theta}{2} = 180° + \frac{\theta}{2}
  • The sum is greater than 180° if D is inside the circle, violating the given condition LB + LD = 180°.
• Case 2: D is outside the circle
  • LB + LD = \frac{\beta}{2} + \frac{\alpha - \theta}{2} = \frac{\alpha + \beta}{2} - \frac{\theta}{2} = 180° - \frac{\theta}{2}
  • If D is outside the circle, LB + LD is less than 180°, a contradiction.
• Therefore, D must be on the circle.
• Method 2: If points C and D lie on the same side of segment AB such that LACB = LADB, then points A, B, C, and D are concyclic.
  • 'Same side' means the points are oriented in a way that visually makes sense, not flipped to opposite sides.
  • Proof is similar to Method 1, and involves drawing the circumcircle of triangle ABC.
    • If D is inside this circle, then LADB > LACB, a contradiction.
    • If D is outside this circle, then LADB < LACB, a contradiction.
    • Thus, point D is on the circumcircle of triangle ABC.

Using Cyclic Quadrilaterals

• Cyclic quadrilaterals are most useful for proving that angles are supplementary or equal.
• Problems usually involve showing that a quadrilateral is cyclic, then using equal angles formed by diagonals and sides to show a pair of angles are equal.
• Although they can show angles are supplementary, usually showing a quadrilateral is cyclic serves to then show that angles are equal.
• Cyclic quadrilaterals can be used to show angles are equal when angles A and B have sides intersecting at C and D, so showing LA = LB involves proving ABCD is a cyclic quadrilateral.
• If A and B are on opposite sides of CD, cyclic quadrilaterals cannot be used to show LA = LB.