angle bcd is a circumscribed angle of circle a. angle bca measures 40°. what is the measure of minor arc bd…

Explanation:

Step1: Recall tangent - radius property

Since $\angle BCD$ is a circumscribed angle and $AB$ and $AD$ are radii and $BC$ and $CD$ are tangents to the circle $A$, $\angle ABC = \angle ADC=90^{\circ}$. Also, $\angle BCA = 40^{\circ}$, so $\angle DCA=40^{\circ}$ (because the two - part circumscribed angle is symmetric about the line from the external point $C$ to the center $A$).

Step2: Find $\angle BAD$

In quadrilateral $ABCD$, the sum of interior angles is $360^{\circ}$. We know $\angle ABC = 90^{\circ}$, $\angle ADC = 90^{\circ}$, and $\angle BCD=\angle BCA+\angle DCA = 80^{\circ}$. Let $\angle BAD=x$. Then $90^{\circ}+90^{\circ}+80^{\circ}+x = 360^{\circ}$. Solving for $x$ gives $x = 100^{\circ}$.

Step3: Use central - arc relationship

The measure of an arc is equal to the measure of its central angle. The central angle of minor arc $BD$ is $\angle BAD$. So the measure of minor arc $BD$ is $100^{\circ}$.

Answer:

$100^{\circ}$