Question

which equation is correct regarding the measure of ∠1?
○ $m∠1 = \frac{1}{2}(a - c)$
○ $m∠1 = \frac{1}{2}(a + c)$
○ $m∠1 = \frac{1}{2}(b - d)$
○ $m∠1 = \frac{1}{2}(b + d)$

Explanation:

Step1: Recall external angle theorem

The measure of an angle formed by two secants intersecting outside a circle is half the difference of the measures of the intercepted arcs. The intercepted arcs are the larger arc \(c\) and the smaller arc \(a\).

Step2: Apply the theorem to \(\angle1\)

The formula is \(m\angle1 = \frac{1}{2}(\text{measure of larger arc} - \text{measure of smaller arc})\). Substitute the arcs: \(m\angle1 = \frac{1}{2}(c - a)\) is not an option, but reordering (since the difference is positive, we take the larger minus smaller, and option 1 uses \(a-c\) but the standard is larger minus smaller, but here the arcs intercepted are the far arc \(c\) and near arc \(a\), so the correct form is \(m\angle1=\frac{1}{2}(c - a)\), but the first option is written as \(m\angle1=\frac{1}{2}(a - c)\) which is equivalent to the negative, but since we take absolute value for angle measure, the correct equation matching the theorem structure with the given arcs is \(m\angle1=\frac{1}{2}(c - a)\), but the only option following the "half the difference of intercepted arcs" is \(m\angle1=\frac{1}{2}(a - c)\) if we consider the labeling, actually the correct theorem is angle outside = 1/2 (major intercepted arc - minor intercepted arc). Here, the major arc intercepted is \(c\), minor is \(a\), so \(m\angle1=\frac{1}{2}(c - a)\), but the first option is \(m\angle1=\frac{1}{2}(a - c)\) which is the negative, but since angle measure is positive, the correct option is the one that represents half the difference of the intercepted arcs, which is \(m\angle1=\frac{1}{2}(a - c)\) (taking the absolute value, the formula is often written as half the difference of the two arcs, regardless of order as we take positive).

Answer:

\(m\angle1 = \frac{1}{2}(a - c)\)