Q1.
Exercise: A cyclic quadrilateral has angles measuring ∠A = 80°, ∠B = 110°, ∠C = 100°, and ∠D = 70°. Can such a quadrilateral be drawn? Explain why or why not.
Answer
Yes, such a quadrilateral can be drawn. Both tests are satisfied.
Test 1 — angle sum of a quadrilateral:
80° + 110° + 100° + 70° = 360° ✓
Test 2 — opposite angles of a cyclic quadrilateral:
∠A + ∠C = 80° + 100° = 180° ✓
∠B + ∠D = 110° + 70° = 180° ✓
80° + 110° + 100° + 70° = 360° ✓
Test 2 — opposite angles of a cyclic quadrilateral:
∠A + ∠C = 80° + 100° = 180° ✓
∠B + ∠D = 110° + 70° = 180° ✓
Both pairs of opposite angles add to 180°, so by Theorem 12 the four vertices must lie on a circle. Such a cyclic quadrilateral exists.
Why the second test is the decisive one: every quadrilateral, cyclic or not, has angle sum 360°, so Test 1 alone proves nothing about the circle. It is Theorem 12 — opposite angles supplementary ⇒ concyclic — that does the real work. Notice also that once ∠A + ∠C = 180° holds, ∠B + ∠D = 360° − 180° = 180° follows automatically; the two conditions are not independent.
Try This: change ∠C to 95° and keep the rest so the sum stays 360° (take ∠D = 75°). Now ∠A + ∠C = 175° ≠ 180°, so a quadrilateral with those angles exists — but no circle can pass through all four of its vertices.