Solution:
Given:
O is the center of the circle and ABCD is a cyclic quadrilateral.
To prove:
i. ∠ABC + ∠ADC = 180°
ii. ∠BAD + ∠BCD = 180°
Construction: Radii OA and OC are drawn
Proof:
Statement | Reasons | |
1. | ∠ABC = \(\frac12\) ∠AOC | ∠ABC is a circumference angle and ∠AOC is central angle standing on same arc ADC. |
2. | ∠ADC = \(\frac12\) reflex ∠AOC | Similar as reason 1. |
3. | ∠ABC + ∠ADC = \(\frac12\angle AOC+\frac12reflex\angle AOC\) | From statement 1. and 2. by addition axiom. |
4. | ∠AOC + reflex ∠AOC = 360° | Sum of angles at the center of the circle |
5. | ∠ABC + ∠ADC = \(\frac12\) 360° = 180° | From statements 3 and 4 by substitution axiom. |
6. | ∠ABC + ∠ADC + ∠BAD + ∠BCD = 360° | Sum of the angles of a quadrilateral. |
7. | 180° + ∠BAD + ∠BCD = 360° Or, ∠BAD + ∠BCD = 360° - 180° = 180° |
From statements 5 and 6 by substitution axiom |
8. | ∠ABC + ∠ADC = 180° and, ∠BAD + ∠BCD = 180° |
From statement 5 and 7 |
hence proved