You know that the sum of angles of a quadrilateral is 360°, even for a non-convex quadrilateral. (Recall the proof.) Now consider a self-intersecting quadrilateral ABCD, where AB and CD intersect at point E. Show that ∠A + ∠B + ∠C + ∠D < 360°. Can you construct ABCD such that ∠A + ∠B + ∠C + ∠D = 2°?
Step-by-step solution
Idea: The crossing point E splits the bow-tie into two triangles. Each triangle has 180°, but the two angles at E are not angles of ABCD, so they must be subtracted.
- (Recall: a diagonal splits an ordinary quadrilateral into two triangles, so its angles add to 2 × 180° = 360°.) In self-intersecting ABCD, AB and CD cross at E, giving triangles EBC and EAD. ∠BEC = ∠AED = θ (vertically opposite angles).1 mark
- ∆EBC: ∠B + ∠C + θ = 180°. ∆EAD: ∠A + ∠D + θ = 180°. Adding: ∠A + ∠B + ∠C + ∠D = 360° − 2θ. Since θ > 0, the sum is less than 360°.1½ marks
- For a sum of 2° we need 360° − 2θ = 2°, so θ = 179°. Construction: take E and four points at distance 1 cm from E: B in direction 0°, C in direction 179°, A in direction 180° and D in direction 359°. AB and CD pass through E and ∠BEC = 179°.1 mark
- ∆EBC is isosceles with apex angle 179°, so ∠B = ∠C = 0.5°; likewise ∠A = ∠D = 0.5°. Sum = 2°. So yes, such an ABCD exists (very thin, almost flat).½ mark
Check: Coordinates with radius 1: B(1, 0), C(cos 179°, sin 179°) ≈ (−0.99985, 0.01745), A(−1, 0), D ≈ (0.99985, −0.01745). Computed angles at A, B, C, D are each 0.500°, total 2.000° ✓.
Answer to write in the exam
AB and CD meet at E; ∠BEC = ∠AED = θ (vertically opposite angles)
∆EBC: ∠B + ∠C = 180° − θ; ∆EAD: ∠A + ∠D = 180° − θ
∴ ∠A + ∠B + ∠C + ∠D = 360° − 2θ < 360° (θ > 0)
Sum = 2° ⇔ θ = 179°: EA = EB = EC = ED, ∠BEC = 179° ⇒ ∠A = ∠B = ∠C = ∠D = 0.5°
∴ Yes, ABCD with angle sum 2° can be constructed.
Common mistakes that cost marks
- Counting the angles at E as angles of ABCD. E is not a vertex.
- Writing ∠A + ∠B + ∠C + ∠D = 360° − θ (subtracting only one angle at E). Both triangles have an angle θ at E.
- Saying a 2° sum is impossible because angles “must add to 360°”. That rule is only for non-self-intersecting quadrilaterals.
How this can come in the exam
In a self-intersecting quadrilateral ABCD, sides AB and CD cross at E with ∠AED = 70°. Then ∠A + ∠B + ∠C + ∠D is
- 360°
- 290°
- 220°
- 140°
Show answer
(C) 220°
360° − 2 × 70° = 220°.
Try one yourself
The angles of a self-intersecting quadrilateral ABCD (AB crossing CD at E) add up to 100°. Find the angle between AB and CD at E.
Show answer
360° − 2θ = 100° ⇒ θ = 130° (so the lines cross at 130° and 50°).
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