If a tangent and a chord intersect at a point on a circle, then the measure of each angle formed is one-half the measure of its intercepted arc.
|}
In order to write the proof we need to prove three cases. Let's start with the first one!
In this case, we have a diameter and a tangent line intersecting at point A. Using this information, we will show that the measure of the angle formed is half the measure of its intercepted arc.
Given:& BA is a diameter.
&BA and CA intersect at point A.
Prove:& m∠ BAC= 1/2mAB
Remember that it was suggested that we use the Tangent Line to Circle Theorem (Theorem 10.1) in this case.
Tangent Line to Circle Theorem
In a plane, a line is tangent to a circle if and only if the line is perpendicular to a radius of the circle at its endpoint on the circle.
By this theorem we can conclude that BA is perpendicular to CA at point A.
Therefore, by the definition of perpendicular lines the measure of ∠ BAC is 90^(∘). Next, we will find the measure of AB. Since BA is a diameter, by the definition of a semicircle AB is a semicircle.
Because the measure of a whole circle is 360 ^(∘), we can deduce that the measure of a semicircle is 180^(∘). Consequently, the measure of ∠ BAC is half the measure of AB.
m∠ BAC= 1/2mAB
b In the second case, we will draw a diagram such that the center of the circle is in the interior of ∠ CAB.
Let's write the Given and Prove statements!
Given:& DA is a diameter.
&BA and CA intersect at point A.
Prove:& m∠ BAC= 1/2mADB
Because DA is a diameter, we know that the measure of ∠ CAD is half the measure of AD by the Tangent and Intersected Chord Theorem Case I (Theorem 10.14).
m∠ CAD = 1/2mAD
Next, we will look at ∠ DAB. Notice that it is an inscribed angle and its intercepted arc is DB.
Therefore, by the Inscribed Angle Theorem (Theorem 10.10) the measure of ∠ DAB is half the measure of DB.
m∠ BAD = 1/2mDB
Next, looking at the above diagram, we can see that the measure of ∠ BAC is the sum of m∠ BAD and m∠ CAD by the Angle Addition Postulate (Postulate 1.4).
m∠ BAC = m∠ BAD + m∠ CAD
Since we also found the measures of ∠ BAD and ∠ CAD in terms of their intercepted arcs, by the Substitution Property of Equality we can substitute them into the above equation.
m∠ BAC = 1/2mDB + 1/2mAD
By the Distributive Property, m∠ BAC = 12(mDB + mAD).
Now, let's examine the intercepted arcs!
From here, we can conclude that mDB + mAD=mADB by the Arc Addition Postulate (Postulate 10.1). Finally, we can substitute mADB for mDB + mAD and complete our proof.
m∠ BAC = 12mADB
c For the third case, the center of the circle will be in the exterior of ∠ CAB.
Let's again write the Given and Prove statements!
Given:& DA is a diameter.
&BA and CA intersect at point A.
Prove:& m∠ BAC= 1/2mAB
To prove the third case, we will follow the same steps as we did in Part B. We will first write the measure of ∠ CAD in terms of its intercepted arc using the Tangent and Intersected Chord Theorem Case I (Theorem 10.14).
m∠ DAC = 1/2mABD
Next, we will investigate ∠ DAB. Notice that its intercepted arc is again DB.
Therefore, by the Inscribed Angle Theorem (Theorem 10.10), we can write the measure of ∠ DAB in terms of the measure of DB.
m∠ DAB = 1/2mDB
Looking at the above diagram, we can see that the measure of ∠ BAC is the difference of m∠ DAC and m∠ DAB by the Angle Addition Postulate (Postulate 1.4).
m∠ BAC = m∠ DAC - m∠ DAB
Now, by the Substitution Property of Equality we can substitute the values of m∠ DAC and m∠ DAB into the equation.
m∠ BAC = 1/2mABD - 1/2mDB
By the Distributive Property, m∠ BAC = 12(mABD - mDB).
Now, let's look at the intercepted arcs!
From here we can conclude that mABD - mDB=mAB by the Arc Addition Postulate (Postulate 10.1). Finally, we can substitute mAB for mABD - mDB and complete our proof.
m∠ BAC = 12mAB
