Question

points a, b, c, and d lie on circle m. line segment bd is a diameter.
what is the measure of angle acd?
○ 45.0°
○ 67.5°
○ 112.5°
○ 135.0°

Explanation:

Step1: Identify right triangle BAD

Since BD is the diameter, $\angle BAD = 90^\circ$ (angle inscribed in a semicircle).

Step2: Find $\angle BDA$

Given $\angle BMA = 90^\circ$, and $MA=MB$ (radii), $\triangle BMA$ is isosceles right triangle, so $\angle MBA = 45^\circ$. In $\triangle BAD$, $\angle BDA = 180^\circ - 90^\circ - 45^\circ = 45^\circ$.

Step3: Find $\angle ACD$

$\angle ACD$ and $\angle ABD$ subtend arc AD. First, note $\angle ABD = 45^\circ$. Also, $\angle ACD$ is an inscribed angle. Arc AD is subtended by $\angle ABD = 45^\circ$, so arc AD measures $90^\circ$. Arc CD = arc BC (marked congruent), and total circle is $360^\circ$. Arc BD is $180^\circ$, so arc BC + arc CD + arc AD = $360^\circ - 180^\circ = 180^\circ$. Since arc BC=arc CD, $2\times\text{arc CD} + 90^\circ = 180^\circ$, so arc CD = $45^\circ$. $\angle CAD$ subtends arc CD, so $\angle CAD = 22.5^\circ$. In $\triangle ACD$, $\angle ADC = 45^\circ$, $\angle CAD = 22.5^\circ$, so $\angle ACD = 180^\circ - 45^\circ - 22.5^\circ = 112.5^\circ$.

Answer:

112.5°