Geometry path · G12
Before this lesson: Circles, tangents and power of a point, Quadrilaterals and their diagonals
Your goal: Prove cyclicity or tangency before applying its consequences.
Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.
Names you may know: Ptolemy’s theorem; Pitot’s theorem; cyclic quadrilateral.
Cyclic and tangential quadrilaterals: the key idea
A convex quadrilateral is cyclic exactly when a pair of opposite angles sums to 180°. Equal angles subtending the same segment give another useful cyclicity test. Ptolemy’s theorem for a convex cyclic quadrilateral ABCD is AC·BD=AB·CD+BC·AD; for any convex quadrilateral, the corresponding inequality is AC·BD≤AB·CD+BC·AD. A tangential quadrilateral has an incircle touching all four sides; equal tangent lengths imply AB+CD=BC+DA, and, conversely, this equality is sufficient for a convex quadrilateral to have an incircle. The converse is a separate theorem; the equal-tangent proof below establishes the forward direction. The Simson–Wallace theorem concerns perpendicular projections from a point on a triangle’s circumcircle, which are collinear.
A worked example
A cyclic quadrilateral has ∠A=70°. Find ∠C.
Opposite angles are supplementary, so ∠C=180°−70°=110°.
Your turn: change one thing
A rectangle has sides 3 and 4. Check Ptolemy.
Try this on paper before opening the explanation.
Compare your reasoning
Each diagonal has length 5. Their product is 25, and the opposite-side products sum to 3·3+4·4=25. The rectangle is cyclic, so equality holds.
Methods and connections
A cyclicity decision
If A and D lie on the same side of BC and ∠BAC=∠BDC, then the four distinct points are concyclic, under the usual noncollinear configuration. This is the converse same-segment angle criterion. For points on opposite sides of the chord, use the supplementary-angle version instead. Checking the side of the chord avoids an incorrect angle interpretation.
Simson–Wallace line
For a point P on the circumcircle of triangle ABC, drop perpendiculars to the three side lines AB,BC,CA. Their feet are collinear, the Simson line of P. Side extensions are allowed. The converse also holds for a point whose three distinct perpendicular feet are collinear. This is a further theorem to investigate after cyclic quadrilaterals and directed angle reasoning.
Pitot’s theorem and its converse
For a convex quadrilateral, equality of opposite-side sums characterises the existence of an incircle. Equal tangents prove the necessary direction, as shown in the written task. The converse requires an additional geometric argument; it does not follow merely by reversing each sentence of the tangent-length proof. Reference: A. F. Beardon, The Mathematical Gazette: Pitot’s theorem.
Pause and check
A trap to avoid: Assuming equal opposite side sums alone always prove tangency for an arbitrary quadrilateral.
Practise and adjust the level
Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.
Interactive practice loads here. You can also use the complete question set below.
Write a complete argument
Prove the opposite-side sum condition for a tangential quadrilateral.
Planning hint
List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.
Read the full solution after your attempt
Let the tangent lengths from vertices A,B,C,D to their adjacent contact points be x,y,z,w. Equal tangents from each vertex give AB=x+y,BC=y+z,CD=z+w,DA=w+x. Therefore AB+CD=x+y+z+w=BC+DA.
My proof notebook
Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.
All nine practice questions
Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.
1. Opposite angles of a cyclic quadrilateral sum to what?
Foundation
- 180°
- 90°
- 360°
- 270°
Hint
They subtend complementary arcs of the circle.
Answer and reasoning
180°. Their intercepted arc measures add to 360°, so the angles add to 180°.
2. A tangential quadrilateral has what?
Foundation
- Equal diagonals always
- Four equal sides always
- A circle tangent to all four sides
- All vertices on a circle by definition
Hint
Tangential refers to an incircle.
Answer and reasoning
A circle tangent to all four sides. Every side touches the same interior circle.
3. In a cyclic quadrilateral, ∠A=70°. What is ∠C?
Core
- 140°
- 110°
- 70°
- 90°
Hint
Subtract from 180°.
Answer and reasoning
110°. Opposite angles are supplementary.
4. For a 3-by-4 rectangle, the diagonal product is what?
Core
- 12
- 24
- 49
- 25
Hint
Each diagonal is 5.
Answer and reasoning
25. 5·5=25.
5. Ptolemy’s equality for cyclic ABCD is which?
Stretch
- AB·BC=CD·DA always
- AC·BD=AB·BC
- AC·BD=AB·CD+BC·AD
- AC+BD=AB+CD
Hint
Pair the two diagonals and the two opposite-side products.
Answer and reasoning
AC·BD=AB·CD+BC·AD. This is the standard cyclic quadrilateral identity.
6. For a convex quadrilateral, does AB+CD=BC+DA imply an incircle exists?
Stretch
- Only for squares
- Only for rational side lengths
- Yes, by the converse of Pitot’s theorem
- No, never
Hint
Convexity is part of the converse theorem.
Answer and reasoning
Yes, by the converse of Pitot’s theorem. For a convex quadrilateral the opposite-side sum equality is both necessary and sufficient. The forward equal-tangent proof alone does not prove the converse.
7. A cyclic quadrilateral has its four vertices where?
Foundation
- On one line
- At the circle centre
- On one circle
- On four unrelated circles only
Hint
Cyclic refers to a common circumcircle.
Answer and reasoning
On one circle. All four distinct vertices lie on the same circle.
8. In cyclic ABCD, ∠B=105°. Find ∠D.
Core
- 90°
- 55°
- 75°
- 105°
Hint
Opposite angles are supplementary.
Answer and reasoning
75°. 180°−105°=75°.
9. Does every rectangle have a circumcircle?
Stretch
- Yes
- No
- Only squares
- Only rectangles with integer sides
Hint
Its opposite angles sum to 180°.
Answer and reasoning
Yes. All angles are 90°, so the cyclicity criterion applies.
Choose your next step
Continue to Trigonometry in geometry. If this felt difficult, return to a prerequisite above. Every lesson stays open.
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