7.6
Tangents and Secants
We start with some issues about parallel lines and circles.
- Theorem 7.14:
Parallel secants intercept congruent arcs between them on a circle.
Given:
$\overleftrightarrow{AB}\parallel \overleftrightarrow{CD}$ Prove: $\overparen{AC}\cong\overparen{BD}$ Proof: Draw chord $\overline{BC}$. Since $\overleftrightarrow{AB}\parallel \overleftrightarrow{CD}$, weh have $\angle 1\cong \angle 2$. Since $m\overparen{AC}=2\cdot m\angle 1$ and $m\overparen{BD}=2\cdot m\angle 2$, then we have $m\overparen{AC}=m\overparen{BD}$, so that $\overparen{AC}\cong\overparen{BD}$. |
- Theorem 7.15:
If a secant and tangent to a circle are parallel, they intercept
congruent arcs between them on the circle.
Given:
$\overleftrightarrow{AT}$ is tangent to $\odot P$ at $T$.
$\overleftrightarrow{BC}\parallel \overleftrightarrow{AT}$. Prove: $\overparen{BT}\cong\overparen{CT}$ Proof: We draw diameter $\overline{TQ}$. This diameter is perpendicular to the tangent $\overleftrightarrow{AT}$. Since the tangent and the secant $\overleftrightarrow{BC}$ are parallel, the diameter is perpendicular to chord $\overline{BC}$ as well. But a diameter that is perpendicular to a chord bisects the chord and its arcs. Therefore $m\overparen{BQ}=m\overparen{CQ}$. The two arcs $\overparen{TBQ}$ and $\overparen{TCQ}$ are semicircles. Using the Arc Addition Postulate, we then know that $\overparen{BT}\cong\overparen{CT}$. |
- Theorem 7.16:
Parallel tangents to a circle intercept congruent arcs between them.
Given:
$\overleftrightarrow{AT}$ and $\overleftrightarrow{BQ}$ are tangent to
$\odot P$ at $T$ and $Q$, respectively.
$\overleftrightarrow{AT}\parallel\overleftrightarrow{BQ}$. Prove: $\overparen{TXQ}\cong\overparen{TYQ}$ |
Proof: It is all too tempting to assume (because of the picture) that $\overleftrightarrow{PT}$ and $\overleftrightarrow{PQ}$are the same line. They are, but it takes some explaining. $\overleftrightarrow{PT}\perp \overleftrightarrow{AT}$ at $T$. Any line that is perpendicular to $\overleftrightarrow{AT}$ will be perpendicular to $\overleftrightarrow{BQ}$ as well, since $\overleftrightarrow{AT}\parallel\overleftrightarrow{BQ}$. Therefore $\overleftrightarrow{PT}\perp\overleftrightarrow{BQ}$. We know also that $\overleftrightarrow{PQ}\perp\overleftrightarrow{BQ}$, since this is the radius drawn to the point of tangency. There is only one perpendicular to $\overleftrightarrow{BQ}$ drawn through point $P$, so $\overleftrightarrow{PT}$ and $\overleftrightarrow{PQ}$ must be the same line. Then $\overparen{TXQ} and $\overparen{TYQ} are both semicircles, so they are congruent.
All of this leads to the theorem about the measure of an angle formed by a tangent and a chord.
- Theorem 7.17:
The Tangent-Chord Theorem
The measure of an angle formed by a tangent segment and a chord drawn to the point of tangency is half the measure of the intercepted arc.
Given:
$\overline{AT}$ is tangent to the circle at $T$. Prove: $m\angle ATB=\dfrac{1}{2}m\overparen{TB}$ Analysis: There are three cases: $\angle ATB$ is right, $\angle ATB$ is acute, and $\angle ATB$ is obtuse. We will deal with the first two, and leave the third for the exercises. |
Case 1: If $\angle ATB$ is a right angle, then $\overline{TB}$ is a diameter and arc $\overparen{TB}$ is a semicircle. The measure of the angle is 90 while the measure of the arc is 180. The measure of the angle is, indeed, half the measure of the intercepted arc.
Case 2: If $\angle ATB$ is acute, then draw chord $\overline{BC}$ parallel to tangent $\overleftrightarrow{AT}$. $m\angle 1=m\angle 2$; $m\angle 2=\frac{1}{2}m\overparen{TC}$; $m\overparen{TC}=m\overparen{TB}$ (since $\overline{BC}\parallel\overleftrightarrow{AT}$), so $\frac{1}{2}m\overparen{TC}=\frac{1}{2}m\overparen{TB}$. Then $m\angle ATB=\frac{1}{2}m\overparen{TB}$. |
We have dealt with angles formed by lines that intersect inside a circle and by (special) lines that intersect on a circle. We now consider lines that intersect outside a circle.
- Theorem 7.18:
The measure of an angle formed by two secants, a secant and a tangent,
or two tangents intersecting at a point outside a circle is half the
difference of the intercepted arcs.
Prove: $m\angle A=\frac{1}{2}\left(m\overparen{CE}-m\overparen{BD}\right)$
Proof: By the Exterior Angle Theorem, $m\angle 2=m\angle 1+m\angle A$. Since $\angle 1$ and $angle 2$ are inscribed angles, their measures are half the measures of the intercepted arcs. That gives $$\frac{1}{2}m\overparen{CE}=\frac{1}{2}m\overparen{BD}+m\angle A.$$ Thus, $$m\angle A=\frac{1}{2}m\overparen{CE}-\frac{1}{2}m\overparen{BD}=\frac{1}{2}\left(m\overparen{CE}=m\overparen{BD}\right).$$
The other two cases are proved in essentially the same way, making use of the Tangent-Chord Theorem.
Example: In the figure, $\overline{AB}$ is tangent to $\odot P$. Arcs $\overparen{BF}$, $\overparen{FE}$, and $\overparen{ED}$ have the measures shown. Determine $m\angle BAF$ and $m\angle CAD$.
Solution: Since
$\overparen{FBC}$ is a semicircle, $m\overparen{BC}=180-40=140$.
Then $$m\angle BAF=\frac{1}{2}(140-40)=50.$$ In the same way, $\overparen{FEDC}$ is a semicircle, so $m\overparen{CD}=180-(24+60)=96$. Then $$m\angle CAD=\frac{1}{2}(96-24)=36.$$ |