Given: Quadrilateral [tex]\(ABCD\)[/tex] inscribed in a circle.

Prove: [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary, [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.

Let the measure of [tex]\(\overline{BCD} = a^\circ\)[/tex]. Because [tex]\(\overline{BCD}\)[/tex] and [tex]\(\overline{BAD}\)[/tex] form a circle, and a circle measures [tex]\(360^\circ\)[/tex], the measure of [tex]\(\overline{BAD}\)[/tex] is [tex]\(360^\circ - a^\circ\)[/tex].

By the inscribed angle theorem, [tex]\(m \angle A = \frac{a}{2}\)[/tex] degrees and [tex]\(m \angle C = \frac{360 - a}{2}\)[/tex] degrees. The sum of the measures of angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex] is [tex]\(\left(\frac{a}{2} + \frac{360 - a}{2}\right)\)[/tex] degrees, which is equal to [tex]\(\frac{360^\circ}{2}\)[/tex], or [tex]\(180^\circ\)[/tex].

Therefore, angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex] are supplementary because their measures add up to [tex]\(180^\circ\)[/tex]. Angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are supplementary because the sum of their measures also equals [tex]\(180^\circ\)[/tex].



Answer :

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.

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