To prove that in a quadrilateral [tex]\(ABCD\)[/tex] inscribed in a circle, the opposite angles are supplementary, we rely on a fundamental property of cyclic quadrilaterals. Specifically, if a quadrilateral is inscribed in a circle, opposite angles of the quadrilateral are supplementary. Here's the detailed, step-by-step solution:
### Given:
- Quadrilateral [tex]\(ABCD\)[/tex] is inscribed in a circle.
- We need to prove that [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary, and [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.
### Solution:
1. Identify Arcs and Corresponding Angles:
- Let the measure of the arc [tex]\( \overset{\frown}{BCD} \)[/tex] be [tex]\(a^\circ\)[/tex].
- Since the full circle is [tex]\(360^\circ\)[/tex], the measure of the arc opposite to [tex]\( \overset{\frown}{BCD} \)[/tex], which is [tex]\( \overset{\frown}{BAD} \)[/tex], will be [tex]\(360^\circ - a^\circ\)[/tex].
2. Relate Arcs to Angles:
- By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of its intercepted arc.
- Therefore, [tex]\(m\angle A = \frac{a}{2}^\circ\)[/tex] because [tex]\(\angle A\)[/tex] intercepts the arc [tex]\( \overset{\frown}{BCD} \)[/tex].
- Similarly, [tex]\(m\angle C = \frac{360 - a}{2}^\circ\)[/tex] because [tex]\(\angle C\)[/tex] intercepts the arc [tex]\( \overset{\frown}{BAD} \)[/tex].
3. Sum of Angles [tex]\( \angle A \)[/tex] and [tex]\( \angle C \)[/tex]:
- To find the sum of angles [tex]\( \angle A \)[/tex] and [tex]\( \angle C \)[/tex]:
[tex]\[
m\angle A + m\angle C = \frac{a}{2} + \frac{360 - a}{2}
\][/tex]
- Combine the terms within the fraction:
[tex]\[
m\angle A + m\angle C = \frac{a + 360 - a}{2} = \frac{360}{2} = 180^\circ
\][/tex]
- Hence, [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary.
4. Sum of Angles [tex]\( \angle B \)[/tex] and [tex]\( \angle D \)[/tex]:
- By similar reasoning, let [tex]\(b^\circ\)[/tex] be the measure of arc [tex]\( \overset{\frown}{CDA}\)[/tex].
- Then, the measure of arc [tex]\( \overset{\frown}{CBA} = 360^\circ - b^\circ\)[/tex].
- Hence, [tex]\(m\angle B = \frac{b}{2}^\circ\)[/tex] and [tex]\(m\angle D = \frac{360 - b}{2}^\circ\)[/tex].
- The sum of angles [tex]\( \angle B \)[/tex] and [tex]\( \angle D \)[/tex]:
[tex]\[
m\angle B + m\angle D = \frac{b}{2} + \frac{360 - b}{2}
\][/tex]
- Combine the terms within the fraction:
[tex]\[
m\angle B + m\angle D = \frac{b + 360 - b}{2} = \frac{360}{2} = 180^\circ
\][/tex]
- Hence, [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.
### Conclusion:
Thus, in any quadrilateral inscribed in a circle, the opposite angles sum up to [tex]\(180^\circ\)[/tex], proving that they are supplementary. In other words, [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary, and [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.
