Ceva in Circumscribed Quadrilateral

Let ABCD be a quadrilateral circumscribed around a circle. Denote the lengths of tangents from the vertices A, B, C, and D to the circle as a, b, c, d, respectively. Finally, let P be the point of intersection of the diagonals AC and BD. Then we have

Proof

Let E be the point of tangency of the incircle on side AB and F be the point of tangency on side BC.

By Brianchon's theorem the lines AF, BP, and CE concur at, say, point Q. Apply Ceva's theorem to ΔABC:

In other words,

And, finally,

Remark

Darij Grinberg has gracefully noted that the theorem has been established by more elementary means elsewhere.

Menelaus and Ceva

• The Menelaus Theorem
• Menelaus Theorem: proofs ugly and elegant - A. Einstein's view
• Ceva's Theorem
• Ceva in Circumscribed Quadrilateral
• Ceva's Theorem: A Matter of Appreciation
• Ceva and Menelaus Meet on the Roads
• Menelaus From Ceva
• Menelaus and Ceva Theorems
• Ceva and Menelaus Theorems for Angle Bisectors
• Ceva's Theorem: Proof Without Words
• Cevian Cradle
• Cevian Nest
• Cevian Triangle
• An Application of Ceva's Theorem
• Trigonometric Form of Ceva's Theorem
• Two Proofs of Menelaus Theorem
• Simultaneous Generalization of the Theorems of Ceva and Menelaus
  • Theorems of Ceva and Menelaus, an Illustrated Generalization
• Menelaus from 3D
• Terquem's Theorem
• Cross Points in a Polygon
• Two Cevians and Proportions in a Triangle, II
  

  References

  1. R. Honsberger, More Mathematical Morsels, MAA, 1991, p. 61.

  

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