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°LA + LC = 180°
    • LB+LD=180°LB + LD = 180°
    • Proof: Angles A and C are inscribed angles.
      • LA+LC=BCD2+BAD2=BCD+BAD2=360°2=180°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=LACDLABD = LACD
      • LBAC=LBDCLBAC = LBDC
      • LACB=LADBLACB = LADB
      • LCBD=LCADLCBD = 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°LB + LD = 180°.
    • Draw the circumcircle of triangleABCtriangle 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=θEF = \theta, AEC=αAEC = \alpha, and ABC=βABC = \beta.
    • Since AEC and ABC make up the whole circle, alpha+beta=360°alpha + beta = 360°.
    • LB+LD=β2+α+θ2=β+α2+θ2=180°+θ2LB + 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°180° if D is inside the circle, violating the given condition LB+LD=180°LB + LD = 180°.
  • Case 2: D is outside the circle
    • LB+LD=β2+α−θ2=α+β2−θ2=180°−θ2LB + 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+LDLB + LD is less than 180°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=LADBLACB = 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 triangleABCtriangle 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 triangleABCtriangle 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=LBLA = 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=LBLA = LB.