Question

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given: a circle with inscribed quadrilateral abcd. prove: ∠a and ∠c are supplementary. 1. let m∠dcb = a°. then m∠dab = 360 - a. 2. by the inscribed - angle theorem, m∠a = a/2. 3. also by the inscribed - angle theorem, m∠c=(360 - a)/2. 4. therefore, m∠a + m∠c = a/2+(360 - a)/2, which simplifies to (choices: 90, 180, 360). check

Explanation:

Step1: Set up angle - measures

Let \(m\angle DCB=a^{\circ}\), then the measure of the arc \(DAB\) is \(360 - a\) (since the sum of the measures of the arcs of a circle is \(360^{\circ}\)).

Step2: Apply inscribed - angle theorem for \(\angle A\)

The measure of an inscribed angle is half the measure of its intercepted arc. So \(m\angle A=\frac{a}{2}\) (as \(\angle A\) intercepts arc \(DCB\)).

Step3: Apply inscribed - angle theorem for \(\angle C\)

\(\angle C\) intercepts arc \(DAB\). So \(m\angle C=\frac{360 - a}{2}\).

Step4: Calculate the sum of \(m\angle A\) and \(m\angle C\)

\[

$$\begin{align*} m\angle A+m\angle C&=\frac{a}{2}+\frac{360 - a}{2}\\ &=\frac{a+(360 - a)}{2}\\ &=\frac{a + 360 - a}{2}\\ &=\frac{360}{2}\\ & = 180 \end{align*}$$

\]

Answer:

180