Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).
Answer: (The question’s ∠CDE is the exterior angle at D when E is on AD produced; below, E is taken on CD produced, so the same exterior angle is called ∠ADE.) With E on CD produced beyond D, ∠ADE + ∠ADC = 180° (linear pair) and ∠ABC + ∠ADC = 180° (opposite angles of a cyclic quadrilateral). So the exterior angle ∠ADE = ∠ABC, the interior opposite angle.
Step-by-step solution
Given: ABCD is cyclic; Side CD is produced beyond D to E
To find: Show that the exterior angle at D equals ∠ABC
To find: Show that the exterior angle at D equals ∠ABC
Idea: Both the exterior angle and ∠ABC are supplements of the same angle ∠ADC: one because of the straight line, the other because ABCD is cyclic.
- C, D, E lie on one line, so the exterior angle at D and the interior angle ∠ADC form a linear pair: ∠ADE + ∠ADC = 180°.½ mark
- ABCD is cyclic, so ∠ABC + ∠ADC = 180° (opposite angles).½ mark
- Both ∠ADE and ∠ABC equal 180° − ∠ADC, so ∠ADE = ∠ABC. The same argument works at every vertex.1 mark
- Note on the example in the question: if E is on CD produced beyond D, then ∠CDE is a straight angle, so the exterior angle at D is ∠ADE. The angle ∠CDE is the exterior angle at D when E is taken on AD produced beyond D. Either way the exterior angle at D is 180° − ∠ADC = ∠ABC.
The exterior angle at D and the interior opposite angle ∠ABC are both supplementary to ∠ADC, so they are equal. The same holds at every vertex of a cyclic quadrilateral.
Answer to write in the exam
C, D, E collinear ⇒ ∠ADE + ∠ADC = 180° (linear pair)
∠ABC + ∠ADC = 180° (opposite angles of a cyclic quadrilateral)
⇒ ∠ADE = 180° − ∠ADC = ∠ABC
∴ Exterior angle at D = interior opposite angle ∠ABC
Common mistakes that cost marks
- Matching the exterior angle with the adjacent interior angle (∠ADC) instead of the opposite one (∠ABC).
- Forgetting to give both reasons: linear pair, and opposite angles of a cyclic quadrilateral.
- Using the result for a quadrilateral that is not cyclic, where it fails.
How this can come in the exam
MCQ (1 mark)
ABCD is a cyclic quadrilateral and side CD is produced to E. If ∠ABC = 95°, the exterior angle ∠ADE is
- 85°
- 95°
- 190°
- 47.5°
Show answer
(B) 95°
The exterior angle of a cyclic quadrilateral equals the interior opposite angle: 95°.
Try one yourself
In a cyclic quadrilateral PQRS, side RS is produced to T, and ∠PST = 72°. Find ∠PQR and ∠PSR.
Show answer
∠PQR = ∠PST = 72° (exterior angle = interior opposite angle); ∠PSR = 180° − 72° = 108°.
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