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Sagot :
Certainly! Here’s a detailed, step-by-step proof of why [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary, and why [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary, for a quadrilateral [tex]\(ABCD\)[/tex] inscribed in a circle.
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### Step-by-Step Proof:
Given:
Quadrilateral [tex]\(ABCD\)[/tex] is inscribed in a circle.
To Prove:
1. [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary.
2. [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.
### Proof for [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex]:
1. Arc Measure Relationship:
- Let the measure of arc [tex]\(BCD\)[/tex] be [tex]\(a^\circ\)[/tex].
- Since [tex]\(BCD\)[/tex] and [tex]\(BAD\)[/tex] form the entire circle, the measure of arc [tex]\(BAD\)[/tex] is [tex]\(360^\circ - a^\circ\)[/tex].
2. Angle-Arc Relationship:
- From the inscribed angle theorem, the measure of an inscribed angle is half of the measure of the intercepted arc.
- Therefore, [tex]\(m \angle A = \frac{1}{2} \times \text{(measure of arc } BCD) = \frac{a}{2}\)[/tex].
- Similarly, [tex]\(m \angle C = \frac{1}{2} \times \text{(measure of arc } BAD) = \frac{360^\circ - a}{2}\)[/tex].
3. Sum of Angles A and C:
- Add [tex]\(m \angle A\)[/tex] and [tex]\(m \angle C\)[/tex]:
[tex]\[ m \angle A + m \angle C = \frac{a}{2} + \frac{360^\circ - a}{2} \][/tex]
- Simplify the expression:
[tex]\[ m \angle A + m \angle C = \frac{a + 360^\circ - a}{2} = \frac{360^\circ}{2} = 180^\circ \][/tex]
4. Conclusion:
- Since the sum of [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] is [tex]\(180^\circ\)[/tex], [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary.
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### Proof for [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex]:
1. Sum of Angles in a Quadrilateral:
- The sum of the interior angles of a quadrilateral is always [tex]\(360^\circ\)[/tex]:
[tex]\[ m \angle A + m \angle B + m \angle C + m \angle D = 360^\circ \][/tex]
2. Use Supplementary Angles:
- From the previous proof, we know [tex]\(m \angle A + m \angle C = 180^\circ\)[/tex].
3. Determine Remaining Sum:
- Therefore:
[tex]\[ m \angle B + m \angle D = 360^\circ - (m \angle A + m \angle C) = 360^\circ - 180^\circ = 180^\circ \][/tex]
4. Conclusion:
- Since the sum of [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] is [tex]\(180^\circ\)[/tex], [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.
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Thus, we have successfully proved that in a cyclic quadrilateral [tex]\(ABCD\)[/tex] inscribed in a circle, [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary, and [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.
---
### Step-by-Step Proof:
Given:
Quadrilateral [tex]\(ABCD\)[/tex] is inscribed in a circle.
To Prove:
1. [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary.
2. [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.
### Proof for [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex]:
1. Arc Measure Relationship:
- Let the measure of arc [tex]\(BCD\)[/tex] be [tex]\(a^\circ\)[/tex].
- Since [tex]\(BCD\)[/tex] and [tex]\(BAD\)[/tex] form the entire circle, the measure of arc [tex]\(BAD\)[/tex] is [tex]\(360^\circ - a^\circ\)[/tex].
2. Angle-Arc Relationship:
- From the inscribed angle theorem, the measure of an inscribed angle is half of the measure of the intercepted arc.
- Therefore, [tex]\(m \angle A = \frac{1}{2} \times \text{(measure of arc } BCD) = \frac{a}{2}\)[/tex].
- Similarly, [tex]\(m \angle C = \frac{1}{2} \times \text{(measure of arc } BAD) = \frac{360^\circ - a}{2}\)[/tex].
3. Sum of Angles A and C:
- Add [tex]\(m \angle A\)[/tex] and [tex]\(m \angle C\)[/tex]:
[tex]\[ m \angle A + m \angle C = \frac{a}{2} + \frac{360^\circ - a}{2} \][/tex]
- Simplify the expression:
[tex]\[ m \angle A + m \angle C = \frac{a + 360^\circ - a}{2} = \frac{360^\circ}{2} = 180^\circ \][/tex]
4. Conclusion:
- Since the sum of [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] is [tex]\(180^\circ\)[/tex], [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary.
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### Proof for [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex]:
1. Sum of Angles in a Quadrilateral:
- The sum of the interior angles of a quadrilateral is always [tex]\(360^\circ\)[/tex]:
[tex]\[ m \angle A + m \angle B + m \angle C + m \angle D = 360^\circ \][/tex]
2. Use Supplementary Angles:
- From the previous proof, we know [tex]\(m \angle A + m \angle C = 180^\circ\)[/tex].
3. Determine Remaining Sum:
- Therefore:
[tex]\[ m \angle B + m \angle D = 360^\circ - (m \angle A + m \angle C) = 360^\circ - 180^\circ = 180^\circ \][/tex]
4. Conclusion:
- Since the sum of [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] is [tex]\(180^\circ\)[/tex], [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary.
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Thus, we have successfully proved that in a cyclic quadrilateral [tex]\(ABCD\)[/tex] inscribed in a circle, [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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