Answer :
Given: quadrilateral [tex]\(ABCD\)[/tex] inscribed in a circle
To prove: [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary, and [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary
1. Understanding the Inscribed Angles:
- Let the measure of arc [tex]\(\overline{BCD}\)[/tex] be [tex]\(a\)[/tex] degrees.
- Since [tex]\(\overline{BCD}\)[/tex] and [tex]\(\overline{BAD}\)[/tex] together form the entire circle (which is [tex]\(360^\circ\)[/tex]), the measure of arc [tex]\(\overline{BAD}\)[/tex] is [tex]\(360^\circ - a\)[/tex].
2. Using the Inscribed Angle Theorem:
- The measure of an inscribed angle is half the measure of its intercepted arc.
- Therefore, the measure of [tex]\(\angle A\)[/tex], which intercepts arc [tex]\(\overline{BCD}\)[/tex], is:
[tex]\[ m\angle A = \frac{a}{2} \text{ degrees} \][/tex]
- The measure of [tex]\(\angle C\)[/tex], which intercepts arc [tex]\(\overline{BAD}\)[/tex], is:
[tex]\[ m\angle C = \frac{360^\circ - a}{2} \text{ degrees} \][/tex]
3. Sum of the Measures of Angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
- The sum of the measures of angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex] is:
[tex]\[ m\angle A + m\angle C = \frac{a}{2} + \frac{360^\circ - a}{2} \][/tex]
- Simplifying this sum, we get:
[tex]\[ m\angle A + m\angle C = \frac{a + 360^\circ - a}{2} = \frac{360^\circ}{2} = 180^\circ \][/tex]
- Therefore, angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex] are supplementary because their measures add up to [tex]\(180^\circ\)[/tex].
4. Similarly for Angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex]:
- Let the measure of arc [tex]\(\overline{ABC}\)[/tex] be [tex]\(b\)[/tex] degrees.
- Similarly, arc [tex]\(\overline{CDA}\)[/tex] will be [tex]\(360^\circ - b\)[/tex].
- The measure of [tex]\(\angle B\)[/tex], which intercepts arc [tex]\(\overline{CDA}\)[/tex], is:
[tex]\[ m\angle B = \frac{b}{2} \text{ degrees} \][/tex]
- The measure of [tex]\(\angle D\)[/tex], which intercepts arc [tex]\(\overline{ABC}\)[/tex], is:
[tex]\[ m\angle D = \frac{360^\circ - b}{2} \text{ degrees} \][/tex]
5. Sum of the Measures of Angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex]:
- The sum of the measures of angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] is:
[tex]\[ m\angle B + m\angle D = \frac{b}{2} + \frac{360^\circ - b}{2} \][/tex]
- Simplifying this sum, we get:
[tex]\[ m\angle B + m\angle D = \frac{b + 360^\circ - b}{2} = \frac{360^\circ}{2} = 180^\circ \][/tex]
- Therefore, angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are supplementary because their measures add up to [tex]\(180^\circ\)[/tex].
In conclusion, for a quadrilateral inscribed in a circle, opposite angles are supplementary. Therefore, [tex]\(\angle A + \angle C = 180^\circ\)[/tex] and [tex]\(\angle B + \angle D = 180^\circ\)[/tex].
To prove: [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary, and [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary
1. Understanding the Inscribed Angles:
- Let the measure of arc [tex]\(\overline{BCD}\)[/tex] be [tex]\(a\)[/tex] degrees.
- Since [tex]\(\overline{BCD}\)[/tex] and [tex]\(\overline{BAD}\)[/tex] together form the entire circle (which is [tex]\(360^\circ\)[/tex]), the measure of arc [tex]\(\overline{BAD}\)[/tex] is [tex]\(360^\circ - a\)[/tex].
2. Using the Inscribed Angle Theorem:
- The measure of an inscribed angle is half the measure of its intercepted arc.
- Therefore, the measure of [tex]\(\angle A\)[/tex], which intercepts arc [tex]\(\overline{BCD}\)[/tex], is:
[tex]\[ m\angle A = \frac{a}{2} \text{ degrees} \][/tex]
- The measure of [tex]\(\angle C\)[/tex], which intercepts arc [tex]\(\overline{BAD}\)[/tex], is:
[tex]\[ m\angle C = \frac{360^\circ - a}{2} \text{ degrees} \][/tex]
3. Sum of the Measures of Angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
- The sum of the measures of angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex] is:
[tex]\[ m\angle A + m\angle C = \frac{a}{2} + \frac{360^\circ - a}{2} \][/tex]
- Simplifying this sum, we get:
[tex]\[ m\angle A + m\angle C = \frac{a + 360^\circ - a}{2} = \frac{360^\circ}{2} = 180^\circ \][/tex]
- Therefore, angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex] are supplementary because their measures add up to [tex]\(180^\circ\)[/tex].
4. Similarly for Angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex]:
- Let the measure of arc [tex]\(\overline{ABC}\)[/tex] be [tex]\(b\)[/tex] degrees.
- Similarly, arc [tex]\(\overline{CDA}\)[/tex] will be [tex]\(360^\circ - b\)[/tex].
- The measure of [tex]\(\angle B\)[/tex], which intercepts arc [tex]\(\overline{CDA}\)[/tex], is:
[tex]\[ m\angle B = \frac{b}{2} \text{ degrees} \][/tex]
- The measure of [tex]\(\angle D\)[/tex], which intercepts arc [tex]\(\overline{ABC}\)[/tex], is:
[tex]\[ m\angle D = \frac{360^\circ - b}{2} \text{ degrees} \][/tex]
5. Sum of the Measures of Angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex]:
- The sum of the measures of angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] is:
[tex]\[ m\angle B + m\angle D = \frac{b}{2} + \frac{360^\circ - b}{2} \][/tex]
- Simplifying this sum, we get:
[tex]\[ m\angle B + m\angle D = \frac{b + 360^\circ - b}{2} = \frac{360^\circ}{2} = 180^\circ \][/tex]
- Therefore, angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are supplementary because their measures add up to [tex]\(180^\circ\)[/tex].
In conclusion, for a quadrilateral inscribed in a circle, opposite angles are supplementary. Therefore, [tex]\(\angle A + \angle C = 180^\circ\)[/tex] and [tex]\(\angle B + \angle D = 180^\circ\)[/tex].
