Question

angle relationships
angle bcd is a circumscribed angle of circle a. angle bca
measures 40
what is the measure of minor arc bd?
80
100
50
40

Explanation:

Step1: Use property of tangent - radius

Since \(BC\) and \(CD\) are tangents to the circle \(A\), \(AB\perp BC\) and \(AD\perp CD\) (tangent to a circle is perpendicular to the radius at the point of tangency). So, \(\angle ABC=\angle ADC = 90^{\circ}\)

Step2: Use angle - sum property of quadrilateral \(ABCD\)

In quadrilateral \(ABCD\), the sum of interior angles is \(360^{\circ}\). Let \(\angle BAD=x\) (which is the central angle corresponding to arc \(BD\)). We know \(\angle BCD = 40^{\circ}\), \(\angle ABC=90^{\circ}\), \(\angle ADC = 90^{\circ}\)

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Another way: Consider \(\triangle ABC\) and \(\triangle ADC\). Since \(AB = AD\) (radii of the same circle) and \(BC=CD\) (tangents from an external point \(C\) to a circle are equal), \(\triangle ABC\cong\triangle ADC\) (by Hypotenuse - Leg congruence in right - angled triangles). So, \(\angle BAC=\angle DAC\)
In right - triangle \(ABC\), \(\angle BAC=90^{\circ}-\angle BCA\) (because \(\angle ABC = 90^{\circ}\) and \(\angle BAC+\angle BCA+\angle ABC=180^{\circ}\)). Given \(\angle BCA = 40^{\circ}\), then \(\angle BAC=50^{\circ}\)
The central angle \(\angle BAD=\angle BAC+\angle DAC\) and since \(\angle BAC=\angle DAC = 50^{\circ}\), \(\angle BAD = 100^{\circ}\)

Answer:

\(100^{\circ}\)