Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Connect with professionals ready to provide precise answers to your questions on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.
Sagot :
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.
### 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.
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
