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Given: Quadrilateral [tex]$ABCD$[/tex] is inscribed in a circle.

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

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

Using 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 equals [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 the measures of the angles in a quadrilateral is [tex]$360^{\circ}$[/tex]. Therefore, [tex]$m \angle A + m \angle C + m \angle B + m \angle D = 360^{\circ}$[/tex].


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.