in the circle below, a is the center, $\overline{bd}$ is a diameter, and $m \angle bac = 130^\circ$. use this information to fill in the blanks.
(a) give a central angle: $\angle \square$
(b) give a semicircle: $\overarc{\square}$
(c) give a major arc: $\overarc{\square}$
(d) find $m \overarc{bc}$ : $\square^\circ$
(e) find $m \overarc{bcd}$ : $\square^\circ$

Question

Accepted Answer

### (a)
# Explanation:
## Step1: Recall central angle definition
A central angle has its vertex at the center of the circle, with sides as radii. Here, $ \angle BAC $ (or others like $ \angle BAD $, $ \angle CAE $ etc.) is a central angle since vertex is $ A $ (center) and sides are radii $ AB, AC $.
# Answer:
$ \angle BAC $ (or $ \angle BAD $, $ \angle CAE $, etc. as valid central angles)

### (b)
# Explanation:
## Step1: Recall semicircle definition
A semicircle is an arc that measures $ 180^\circ $, formed by a diameter. Since $ BD $ is a diameter, arc $ BED $ (or $ BCD $) is a semicircle (passes through half the circle).
# Answer:
$ \overarc{BED} $ (or $ \overarc{BCD} $, as semicircles formed by diameter $ BD $)

### (c)
# Explanation:
## Step1: Recall major arc definition
A major arc is an arc greater than $ 180^\circ $. Arc $ BCD $ (wait, no, major arc could be $ \overarc{BCE} $? Wait, with $ \angle BAC = 130^\circ $, arc $ BC $ is $ 130^\circ $, so major arc $ BDC $ (no, better: major arc $ BEC $? Wait, actually, major arc $ BCD $ is not, wait, major arc should be longer than $ 180^\circ $. Since $ BD $ is diameter ($ 180^\circ $), major arc $ BAC $? No, wait, major arc $ BDC $ – no, let's see: central angle for minor arc $ BC $ is $ 130^\circ $, so major arc $ BC $ would be $ 360 - 130 = 230^\circ $, so $ \overarc{BCD} $ is not, wait, $ \overarc{BCE} $? Wait, actually, major arc $ BDC $ is incorrect. Wait, the major arc corresponding to minor arc $ BC $ is $ \overarc{BDC} $? No, better: major arc $ BEDC $? Wait, maybe $ \overarc{BCD} $ is not, wait, let's re-express. A major arc is an arc that is more than a semicircle. So for example, $ \overarc{BCD} $ – no, $ BD $ is diameter, so $ \overarc{BCD} $ is semicircle? Wait, no, $ BD $ is diameter, so arc $ BCD $ is semicircle? Wait, no, $ B $ to $ D $ through $ C $: if $ \angle BAC = 130^\circ $, then $ \angle CAD = 180 - 130 = 50^\circ $? Wait, no, $ BD $ is diameter, so $ \angle BAD = 180^\circ $? Wait, no, $ A $ is center, $ BD $ is diameter, so $ \angle BAD = 180^\circ $? Wait, the diagram shows $ \angle BAC = 130^\circ $, so $ \angle CAD = 180 - 130 = 50^\circ $? Wait, maybe I messed up. Wait, major arc: an arc that is greater than $ 180^\circ $. So for example, arc $ BEC $ (if $ E $ is another point) or $ \overarc{BCD} $ – no, let's take major arc $ BDC $ is not, wait, the major arc could be $ \overarc{BC D} $ no, wait, let's think again. The minor arc $ BC $ is $ 130^\circ $, so major arc $ BC $ is $ 360 - 130 = 230^\circ $, so $ \overarc{BCD} $ is not, wait, $ \overarc{BEDC} $? No, maybe $ \overarc{BCD} $ is incorrect. Wait, the correct major arc would be $ \overarc{BDC} $ – no, I think the intended answer is $ \overarc{BCD} $ is not, wait, maybe $ \overarc{BCE} $ – no, perhaps the major arc is $ \overarc{BCD} $ (but no, $ BD $ is diameter, so $ \overarc{BCD} $ is semicircle? Wait, no, $ BD $ is diameter, so the arc from $ B $ to $ D $ through $ C $ is $ \overarc{BCD} $, which is $ 130^\circ + 50^\circ = 180^\circ $? Wait, no, $ \angle BAC = 130^\circ $, so arc $ BC = 130^\circ $, $ \angle CAD = 180 - 130 = 50^\circ $, so arc $ CD = 50^\circ $, so arc $ BCD = 130 + 50 = 180^\circ $, which is semicircle. Oh! So I made a mistake. Then major arc should be, for example, $ \overarc{BEC} $ (if $ E $ is a point), but in the diagram, $ E $ is on the circle. Wait, maybe major arc $ BDC $ is not, wait, the major arc corresponding to minor arc $ BC $ is $ \overarc{BDC} $ – no, minor arc $ BC $ is $ 130^\circ $, so major arc $ BC $ is $ 360 - 130 = 230^\circ $, so $ \overarc{BCD} $ is not, wait, $ \overarc{BEDC} $? No, perhaps the major arc is $ \overarc{BCD} $ is incorrect. Wait, let's check the central angle: major arc has central angle greater than $ 180^\circ $. So for example, central angle for major arc $ BCD $ – no, $ BD $ is diameter, central angle $ 180^\circ $. Wait, maybe the major arc is $ \overarc{BCE} $, but I think the intended answer is $ \overarc{BCD} $ is not, wait, maybe $ \overarc{BAC} $ – no. Wait, perhaps the major arc is $ \overarc{BDC} $ – no, I'm confused. Wait, the problem says "Give a major arc" – so any arc greater than $ 180^\circ $. So for example, $ \overarc{BCD} $ is not, but $ \overarc{BED} $ is semicircle. Wait, maybe $ \overarc{BCE} $ – no, let's see: minor arc $ BC = 130^\circ $, so major arc $ BC $ is $ 360 - 130 = 230^\circ $, so $ \overarc{BDC} $ (wait, $ B $ to $ D $ to $ C $ – no, $ D $ to $ C $ is $ 50^\circ $, $ B $ to $ D $ is $ 180^\circ $, so $ B $ to $ D $ to $ C $ is $ 180 + 50 = 230^\circ $, so $ \overarc{BDC} $ is a major arc (230°), so that's a major arc. So $ \overarc{BDC} $ is a major arc.
# Answer:
$ \overarc{BDC} $ (or other valid major arcs like $ \overarc{BEC} $ if applicable, but $ \overarc{BDC} $ is correct as $ 230^\circ > 180^\circ $)

### (d)
# Explanation:
## Step1: Recall central angle - arc relationship
The measure of an arc is equal to the measure of its central angle. $ \angle BAC $ is the central angle for arc $ BC $, and $ m\angle BAC = 130^\circ $, so $ m\overarc{BC} = 130^\circ $.
# Answer:
$ 130 $

### (e)
# Explanation:
## Step1: Recall semicircle measure
A semicircle (arc formed by diameter) measures $ 180^\circ $. Arc $ BCD $ is a semicircle? Wait, no, $ BD $ is diameter, so arc $ BCD $ – wait, $ \overarc{BCD} $ is the arc from $ B $ to $ C $ to $ D $. We know $ m\overarc{BC} = 130^\circ $, and $ \angle CAD $: since $ BD $ is diameter, $ \angle BAD = 180^\circ $, and $ \angle BAC = 130^\circ $, so $ \angle CAD = 180^\circ - 130^\circ = 50^\circ $, so $ m\overarc{CD} = 50^\circ $. Then $ m\overarc{BCD} = m\overarc{BC} + m\overarc{CD} = 130^\circ + 50^\circ = 180^\circ $? Wait, no, that's semicircle. Wait, but $ BD $ is diameter, so arc $ BCD $ (from $ B $ to $ C $ to $ D $) is a semicircle? Wait, no, $ BD $ is diameter, so arc $ BED $ (from $ B $ to $ E $ to $ D $) is semicircle, but $ B $ to $ C $ to $ D $: $ \overarc{BC} = 130^\circ $, $ \overarc{CD} = 50^\circ $, sum to $ 180^\circ $, so yes, $ m\overarc{BCD} = 180^\circ $. Wait, but that's a semicircle. Wait, maybe I made a mistake. Wait, $ BD $ is diameter, so the arc from $ B $ to $ D $ through $ C $ is $ \overarc{BCD} $, which is $ m\overarc{BC} + m\overarc{CD} $. Since $ BD $ is diameter, $ m\overarc{BD} = 180^\circ $, but $ \overarc{BCD} $ is $ \overarc{BC} + \overarc{CD} $. Wait, $ \angle BAC = 130^\circ $ (central angle for $ \overarc{BC} $), and $ \angle CAD $: since $ A $ is center, $ AC $ and $ AD $ are radii, and $ BD $ is diameter, so $ \angle BAD = 180^\circ $, so $ \angle CAD = \angle BAD - \angle BAC = 180 - 130 = 50^\circ $, so $ m\overarc{CD} = 50^\circ $. Thus, $ m\overarc{BCD} = 130 + 50 = 180^\circ $.
# Answer:
$ 180 $

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QUESTION IMAGE

Question

Explanation:

(a)

Step1: Recall central angle definition

Step1: Recall semicircle definition

Step1: Recall major arc definition

Answer:

(b)