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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, let's follow the given steps and numerical results in a structured manner.
Step-by-Step Solution:
1. Define the Measures of Arcs and Angles:
- Given that the measure of [tex]\(\overline{BCD}\)[/tex] is [tex]\(a^{\circ}\)[/tex].
- Since [tex]\(\overline{BCD}\)[/tex] and [tex]\(\overline{BAD}\)[/tex] form a circle, the sum of their measures is [tex]\(360^{\circ}\)[/tex]. Therefore, the measure of [tex]\(\overline{BAD}\)[/tex] is [tex]\(360 - a^{\circ}\)[/tex].
2. Calculate the Measures of Angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
- According to the theorem (Inscribed Angle Theorem), we know:
[tex]\[ m \angle A = \frac{a}{2} \quad \text{degrees} \][/tex]
[tex]\[ m \angle C = \frac{360 - a}{2} \quad \text{degrees} \][/tex]
3. Sum of Angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
- Next, let's find the sum of the measures of angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ m \angle A + m \angle C = \left( \frac{a}{2} + \frac{360 - a}{2} \right) \quad \text{degrees} \][/tex]
- Simplifying the expression inside the parenthesis:
[tex]\[ \frac{a}{2} + \frac{360 - a}{2} = \frac{a + 360 - a}{2} = \frac{360}{2} = 180 \quad \text{degrees} \][/tex]
- Thus, [tex]\(m \angle A + m \angle C = 180^{\circ}\)[/tex]. This proves that angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex] are supplementary because their measures add up to [tex]\(180^{\circ}\)[/tex].
4. Prove Angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are Supplementary:
- The sum of the measures of the angles in a quadrilateral is always [tex]\(360^{\circ}\)[/tex]. Therefore:
[tex]\[ m \angle A + m \angle C + m \angle B + m \angle D = 360^{\circ} \][/tex]
- We have already established that [tex]\(m \angle A + m \angle C = 180^{\circ}\)[/tex]. Substituting into the equation:
[tex]\[ 180^{\circ} + m \angle B + m \angle D = 360^{\circ} \][/tex]
- Solving for the sum of [tex]\(m \angle B\)[/tex] and [tex]\(m \angle D\)[/tex]:
[tex]\[ m \angle B + m \angle D = 360^{\circ} - 180^{\circ} = 180^{\circ} \][/tex]
- Hence, angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are supplementary because their measures add up to [tex]\(180^{\circ}\)[/tex].
Summary:
Using the given measures and calculations, we have shown that [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary (their sum is [tex]\(180^{\circ}\)[/tex]). Likewise, [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary since their measures also add up to [tex]\(180^{\circ}\)[/tex]. Thus, the proof is complete.
Step-by-Step Solution:
1. Define the Measures of Arcs and Angles:
- Given that the measure of [tex]\(\overline{BCD}\)[/tex] is [tex]\(a^{\circ}\)[/tex].
- Since [tex]\(\overline{BCD}\)[/tex] and [tex]\(\overline{BAD}\)[/tex] form a circle, the sum of their measures is [tex]\(360^{\circ}\)[/tex]. Therefore, the measure of [tex]\(\overline{BAD}\)[/tex] is [tex]\(360 - a^{\circ}\)[/tex].
2. Calculate the Measures of Angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
- According to the theorem (Inscribed Angle Theorem), we know:
[tex]\[ m \angle A = \frac{a}{2} \quad \text{degrees} \][/tex]
[tex]\[ m \angle C = \frac{360 - a}{2} \quad \text{degrees} \][/tex]
3. Sum of Angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
- Next, let's find the sum of the measures of angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex]:
[tex]\[ m \angle A + m \angle C = \left( \frac{a}{2} + \frac{360 - a}{2} \right) \quad \text{degrees} \][/tex]
- Simplifying the expression inside the parenthesis:
[tex]\[ \frac{a}{2} + \frac{360 - a}{2} = \frac{a + 360 - a}{2} = \frac{360}{2} = 180 \quad \text{degrees} \][/tex]
- Thus, [tex]\(m \angle A + m \angle C = 180^{\circ}\)[/tex]. This proves that angles [tex]\(A\)[/tex] and [tex]\(C\)[/tex] are supplementary because their measures add up to [tex]\(180^{\circ}\)[/tex].
4. Prove Angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are Supplementary:
- The sum of the measures of the angles in a quadrilateral is always [tex]\(360^{\circ}\)[/tex]. Therefore:
[tex]\[ m \angle A + m \angle C + m \angle B + m \angle D = 360^{\circ} \][/tex]
- We have already established that [tex]\(m \angle A + m \angle C = 180^{\circ}\)[/tex]. Substituting into the equation:
[tex]\[ 180^{\circ} + m \angle B + m \angle D = 360^{\circ} \][/tex]
- Solving for the sum of [tex]\(m \angle B\)[/tex] and [tex]\(m \angle D\)[/tex]:
[tex]\[ m \angle B + m \angle D = 360^{\circ} - 180^{\circ} = 180^{\circ} \][/tex]
- Hence, angles [tex]\(B\)[/tex] and [tex]\(D\)[/tex] are supplementary because their measures add up to [tex]\(180^{\circ}\)[/tex].
Summary:
Using the given measures and calculations, we have shown that [tex]\(\angle A\)[/tex] and [tex]\(\angle C\)[/tex] are supplementary (their sum is [tex]\(180^{\circ}\)[/tex]). Likewise, [tex]\(\angle B\)[/tex] and [tex]\(\angle D\)[/tex] are supplementary since their measures also add up to [tex]\(180^{\circ}\)[/tex]. Thus, the proof is complete.
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