Cyclic Quadrilateral Angle Dispute
In a school geometry project, four points A, B, C and D are marked in order on a circle. One student reports ∠ABC = 95° and ∠BCD = 70°. Another student concludes that ∠CDA = 85°, ∠DAB = 110°, and therefore AD is parallel to BC. The teacher points out that the side lengths AB = 6 cm, BC = 8 cm and diagonal AC = 10 cm must also be respected, and asks the class to test the angle report before accepting the parallelism. Use the converse of the cyclic quadrilateral theorem, the fact that opposite angles of a cyclic quadrilateral are supplementary, and the Pythagoras theorem. The circle is the same for all four points, and the order A-B-C-D around the circle is fixed. No point is assumed to be a centre.
1. Using triangle ABC, determine the correct value of ∠ABC and explain why the reported 95° is impossible.
Read the worked answer
In triangle ABC, AB = 6, BC = 8 and AC = 10. Check: AB² + BC² = 36 + 64 = 100 = AC². By the converse of the Pythagoras theorem, the right angle is opposite AC, so ∠ABC = 90°.
A reported ∠ABC = 95° would require AC² > AB² + BC², i.e. AC > 10 cm. Since AC is exactly 10 cm, 95° is impossible for these side lengths. The angle report is inconsistent with the given triangle.
2. Using the corrected ∠ABC, find ∠CDA and test whether AD is parallel to BC, taking ∠BCD = 70°.
Read the worked answer
Corrected ∠ABC = 90°. In a cyclic quadrilateral, opposite angles are supplementary, so ∠CDA = 180° − 90° = 90°.
Also ∠DAB = 180° − ∠BCD = 180° − 70° = 110°.
If AD were parallel to BC, then with AB as transversal the co-interior angles ∠DAB and ∠ABC would add to 180°. But 110° + 90° = 200° ≠ 180°. Therefore AD is not parallel to BC.
3. Find ∠ACB and hence ∠ACD, using the corrected triangle and ∠BCD = 70°.
Read the worked answer
In right triangle ABC, sin ∠ACB = AB/AC = 6/10 = 0.6, so ∠ACB ≈ 36.87°.
Because A-B-C-D are in cyclic order, diagonal CA lies inside ∠BCD, so ∠BCD = ∠BCA + ∠ACD. Hence ∠ACD = 70° − 36.87° = 33.13° approximately.