How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
Step-by-step solution
Idea: The equal ticks say the four segments from O are radii. Each side of the quadrilateral and two radii make an isosceles triangle with equal base angles, so every angle of the quadrilateral is a sum of two of the letters p, q, u, v.
- The ticks show OA = OB = OC = OD (radii). Isosceles triangles: in ΔOAB, ∠OBA = ∠OAB = p; in ΔOBC, ∠OCB = ∠OBC = q; in ΔOCD, ∠ODC = ∠OCD = u; in ΔODA, ∠OAD = ∠ODA = v.1 mark
- So the angles of ABCD are: ∠A = p + v, ∠B = p + q, ∠C = q + u, ∠D = u + v.½ mark
- Angle sum of the quadrilateral: 2(p + q + u + v) = 360°, so p + q + u + v = 180°.1 mark
- Hence ∠A + ∠C = p + v + q + u = 180° and ∠B + ∠D = p + q + u + v = 180°.½ mark
- (This figure has the centre inside the quadrilateral. If the centre lies outside, one of the four triangles is subtracted instead of added, and the same result follows.)
Answer to write in the exam
OA = OB = OC = OD (radii)
∠OBA = ∠OAB = p, ∠OCB = ∠OBC = q, ∠ODC = ∠OCD = u, ∠OAD = ∠ODA = v (angles opposite equal sides)
∠A = p + v, ∠B = p + q, ∠C = q + u, ∠D = u + v
∠A + ∠B + ∠C + ∠D = 360° ⇒ 2(p + q + u + v) = 360° ⇒ p + q + u + v = 180°
∴ ∠A + ∠C = 180° and ∠B + ∠D = 180°
Common mistakes that cost marks
- Assuming p = q = u = v. Only the two base angles inside each triangle are equal.
- Matching base angles across different triangles (e.g. ∠OBA = q): each equal pair belongs to one isosceles triangle.
- Forgetting to use the quadrilateral angle sum 360°, which is what turns the sum into 180°.
How this can come in the exam
In the figure, p = 30°, q = 40° and u = 50°. Then v is
- 40°
- 50°
- 60°
- 70°
Show answer
(C) 60°
p + q + u + v = 180°, so v = 180° − 120° = 60°.
Try one yourself
In the same figure, p = 25°, q = 35° and v = 55°. Find u, ∠A and ∠C, and check that ∠A + ∠C = 180°.
Show answer
u = 180° − 115° = 65°. ∠A = p + v = 80°, ∠C = q + u = 100°; 80° + 100° = 180° ✓.
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