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 (A) and sides as radii. $\angle BAC$ has vertex A, sides AB and AC (radii), so it's a central angle. Other options: $\angle BAD$, $\angle DAE$, etc. (choosing $\angle BAC$ here).
# Answer:
$\angle BAC$ (or $\angle BAD$, $\angle DAE$, etc.)

### (b)
# Explanation:
## Step1: Recall semicircle definition
A semicircle is an arc measuring $180^\circ$, formed by a diameter. $\overline{BD}$ is a diameter, so arc $\overarc{BCD}$ or $\overarc{BED}$ (using diameter BD, arc from B to D through C or E is a semicircle; $\overarc{BCD}$ is a semicircle here as BD is diameter, so arc BCD spans half the circle).
# Answer:
$\overarc{BCD}$ (or $\overarc{BED}$)

### (c)
# Explanation:
## Step1: Recall major arc definition
A major arc is an arc greater than $180^\circ$. Arc $\overarc{BCD}$ is a semicircle ($180^\circ$), so a major arc could be $\overarc{BCE}$ (but better: $\overarc{BC D}$ is semicircle, so $\overarc{BCD}$ is not major. Wait, major arc is more than $180^\circ$. Let's see: central angle $\angle BAC = 130^\circ$, so arc BC is $130^\circ$. Then arc BDC: no, wait, major arc for BC would be $\overarc{BCD}$? No, wait, major arc is longer than semicircle. So arc $\overarc{BCE}$? Wait, better: major arc from B to C through E or D? Wait, $\overarc{BCD}$ is semicircle (180), so major arc would be $\overarc{BCE}$? No, let's use the diagram. BD is diameter, so arc BCD is 180. A major arc could be $\overarc{BCE}$? Wait, no, let's think: central angle for arc BC is 130, so major arc BC would be $360 - 130 = 230^\circ$, so arc $\overarc{BCD}$? No, wait, arc BC is 130, so major arc BC is $\overarc{B D C}$? Wait, maybe $\overarc{BCE}$ is not. Wait, the major arc should be named with three points. So $\overarc{BCD}$ is semicircle, so major arc could be $\overarc{BCE}$? No, perhaps $\overarc{BC D}$ is not. Wait, maybe $\overarc{BCE}$ is a major arc? Wait, no, let's check: the circle has center A, diameter BD. So arc BCD is 180. A major arc (greater than 180) would be, for example, $\overarc{BCE}$? Wait, maybe the answer is $\overarc{BCD}$? No, that's semicircle. Wait, maybe I made a mistake. Wait, major arc is more than 180, so arc $\overarc{BC D}$ is 180, so no. Wait, the central angle for arc BC is 130, so the major arc BC is $360 - 130 = 230^\circ$, so the major arc would be $\overarc{B D C}$? Wait, the notation: $\overarc{BCD}$ is from B to C to D, which is 180 (semicircle). So major arc BC would be $\overarc{B E C}$? No, maybe the answer is $\overarc{BCD}$ is not, so let's take $\overarc{BCE}$? Wait, perhaps the correct major arc is $\overarc{BCD}$? No, that's semicircle. Wait, maybe the problem allows $\overarc{BCD}$ as a semicircle, so a major arc could be $\overarc{BCE}$? Wait, maybe I should look at the diagram again. The circle has points B, C, D, E. BD is diameter. So arc BCD is 180 (semicircle). Arc BCE: from B to C to E. If arc BC is 130, arc CE: let's see, angle BAD: since BD is diameter, angle BAD is 90? No, A is center, so AB and AD are radii, so triangle ABD is isoceles? Wait, no, BD is diameter, so angle BAD: wait, $\angle BAC = 130^\circ$, so $\angle BAD$: since BD is diameter, the straight line, so $\angle BAC + \angle CAD = 180$? Wait, no, A is center, so AC, AB, AD, AE are radii. So $\angle BAC = 130^\circ$, so arc BC is 130^\circ. Then, since BD is diameter, arc BCD is 180^\circ, so arc CD is $180 - 130 = 50^\circ$? Wait, no, $\angle BAC = 130^\circ$, so arc BC is 130^\circ. Then, BD is diameter, so arc BCD is 180^\circ, so arc CD is $180 - 130 = 50^\circ$? Wait, no, $\angle BAD$: since BD is diameter, $\angle BAD$ is a straight angle? No, A is center, so AB and AD are radii, so BD is diameter, so $\angle BAD$ is a straight angle? Wait, no, in the diagram, A is the center, so AB, AD are radii, so BD is diameter, so $\angle BAD$ is 180^\circ? Wait, no, the diagram shows angle BAC is 130, so AC is a radius, AB is a radius, so triangle BAC is isoceles with AB=AC. Then AD is a radius, so angle CAD: since BD is diameter, angle BAD is 180 - angle BAC? Wait, no, points B, A, D: are they colinear? Yes! Because BD is a diameter, so B, A, D are colinear (center A, so BD is a straight line through A). Oh! Wait, that's the key. BD is a diameter, so B, A, D are colinear (A is center, so BD passes through A, so B---A---D is a straight line). So $\angle BAD = 180^\circ$, because B, A, D are colinear. So $\angle BAC = 130^\circ$, so $\angle CAD = 180^\circ - 130^\circ = 50^\circ$? Wait, no, if B, A, D are colinear (straight line), then $\angle BAC + \angle CAD = 180^\circ$, so $\angle CAD = 180 - 130 = 50^\circ$. So arc CD is 50^\circ (since central angle $\angle CAD = 50^\circ$). Then, arc BC is 130^\circ (central angle $\angle BAC = 130^\circ$), arc CD is 50^\circ (central angle $\angle CAD = 50^\circ$). So a major arc: arc BCD is 180 (semicircle), so major arc would be arc BCE (but E is another point). Wait, major arc BC: $360 - 130 = 230^\circ$, so arc BDC (B to D to C) is 180 + 50 = 230^\circ? Wait, arc BD is 180 (diameter), arc DC is 50, so arc BDC is 180 + 50 = 230^\circ, which is a major arc. So arc $\overarc{BDC}$ is a major arc.
# Answer:
$\overarc{BDC}$ (or other major arcs like $\overarc{BEC}$ if applicable, but $\overarc{BDC}$ is $230^\circ$)

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

### (e)
# Explanation:
## Step1: Recall semicircle measure
A semicircle measures $180^\circ$. $\overarc{BCD}$ is a semicircle (since $\overline{BD}$ is a diameter), so $m\overarc{BCD} = 180^\circ$.
# Answer:
$180^\circ$

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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)