At Westonci.ca, we connect you with experts who provide detailed answers to your most pressing questions. Start exploring now! Join our Q&A platform to get precise answers from experts in diverse fields and enhance your understanding. Join our Q&A platform to connect with experts dedicated to providing accurate answers to your questions in various fields.
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.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.
