Question

given: a circle with inscribed quadrilateral abcd
prove: ∠a and ∠c are supplementary.

1. let ( moverarc{dcb} = a^circ )

then ( moverarc{dab} = 360 - a ).

1. by the inscribed angle theorem, ( mangle a = a/2 ).
2. also by the inscribed angle theorem, ( mangle c = )

options: ( 180 - a/2 ), ( 360 - a/2 ), ( (360 - a)/2 )
(image of a circle with inscribed quadrilateral abcd, vertices labeled a, b, c, d in order on the circle.)

Explanation:

Step1: Recall the total circle measure

The total measure of a circle is \(360^\circ\). If \(m\widehat{DCB}=a^\circ\), then the measure of the arc opposite to it (arc \(DAB\)) is \(360 - a\) degrees (since the sum of arcs around a circle is \(360^\circ\)). But for angle \(C\), we need to find the inscribed angle over its intercepted arc. The intercepted arc for angle \(C\) is arc \(DAB\), which has measure \(360 - a\) degrees. By the inscribed angle theorem, the measure of an inscribed angle is half the measure of its intercepted arc. So \(m\angle C=\frac{360 - a}{2}\), which is \((360 - a)/2\).

Step2: Verify with the theorem

The inscribed angle theorem states that an angle \(\theta\) subtended by an arc at the circumference of a circle is half the central angle subtended by the same arc. For \(\angle C\), the intercepted arc is arc \(DAB\) with measure \(360 - a\) (from step 1, since arc \(DCB\) is \(a\)). So applying the theorem, \(m\angle C=\frac{1}{2}\times(360 - a)=\frac{360 - a}{2}\).

Answer:

\(\frac{360 - a}{2}\) (or \((360 - a)/2\))