12.05 Angles from intersecting chords, secants, and tangents

Angles inside the circle

When two chords (or the secants that contain them) intersect inside a circle, a special relationship exists between the measures of the angles formed and the intercepted arcs.

A chord is a line segment that goes from one point on the circle's circumference to another, in the figure below $\overline{AB}$AB and $\overline{CD}$CD are chords.

 

Exploration

Press the double arrow in the navigation bar of the applet below. Can you explain each step? What's the relationship between $\angle CBF$∠CBF and arcs $CF$CF and $ED$ED?

Let's look at what's happening in each step of the applet.

1.	$\overline{CD}$CDand $\overline{EF}$EFare chords of circle $A$A	Given
2.	Draw $\overline{CE}$CE	Any two points define a line.
3.	Let the measureof arc$ED=a^\circ$ED=a°
	$m\angle ECD=(\frac{a}{2})^\circ$m∠ECD=(a2​)°	Inscribed angle theorem
4.	Let the measure of arc$CF=b^\circ$CF=b°
	$m\angle CEF=(\frac{b}{2})^\circ$m∠CEF=(b2​)°	Inscribed angle theorem
5.	$m\angle CBF=m\angle ECD+m\angle CEF$m∠CBF=m∠ECD+m∠CEF	Exterior angle theorem
6.	$m\angle CBF=(\frac{a}{2}+\frac{b}{2})^\circ$m∠CBF=(a2​+b2​)°	Substitution
	$m\angle CBF=(\frac{a+b}{2})^\circ$m∠CBF=(a+b2​)°	Simplify

 

The relationship we've proven can be summarized in the following theorem:

 

Practice questions

Question 1

Find the value of $x$x.

Reveal SolutionWatch video

Question 2

Angles on the circle

Recall that by the inscribed angle theorem, an inscribed angle is always half the measure of its intercepted arc. This theorem also applies to angles on the circumference of the circle formed by intersecting secants and tangents. 

A secant is a line segment or ray that has one endpoint on the circumference of the circle and the other endpoint outside the circle, so it touches and passes through two points on a circle's circumference. $\overline{AE}$AE and $\overline{AC}$AC are secants.

A tangent is a line that intersects the circumference of a circle in exactly one point, which we call the point of tangency.

A tangent is perpendicular to the radius from the point of tangency. 

Angles outside the circle

There also exists a special relationship when secants and tangents form angles outside the circle.

Exploration

Press the double arrow in the navigation bar of the applet below. Can you explain each step?

What's the relationship between $a$a, $b$b, and $c$c in each of the following cases

1. When $\overleftrightarrow{AB}$›‹AB and $\overleftrightarrow{CD}$›‹CD are both secants to circle $O$O.
2. When $\overleftrightarrow{AP}$›‹AP is tangent to circle $O$O? (Hint: move point $A$A until point $B$B disappears).
3. When both $\overleftrightarrow{AP}$›‹AP and $\overleftrightarrow{AP}$›‹AP are tangent to circle $O$O?

Let's walk through the case where $\overleftrightarrow{AB}$›‹AB and $\overleftrightarrow{CD}$›‹CD are both secants to circle $O$O.

1.	$\overleftrightarrow{AB}$›‹AB and $\overleftrightarrow{CD}$›‹CD are secants to circle $O$O	Given
2.	Draw $\overline{AD}$AD	Any two points define a line.
3.	Let the measureof arc$AC=a^\circ$AC=a°
	$m\angle ADC=(\frac{a}{2})^\circ$m∠ADC=(a2​)°	Inscribed angle theorem
4.	Let the measureof arc$BD=b^\circ$BD=b°
	$m\angle DAP=(\frac{b}{2})^\circ$m∠DAP=(b2​)°	Inscribed angle theorem
5.	Let $m\angle APC=c^\circ$m∠APC=c°
	$m\angle ADC=m\angle APD+m\angle PAD$m∠ADC=m∠APD+m∠PAD	Exterior angle theorem
	$(\frac{a}{2})^\circ=(c+\frac{b}{2})^\circ$(a2​)°=(c+b2​)°	Substitution
	$\frac{a}{2}-\frac{b}{2}=c$a2​−b2​=c	Subtraction property of equality
	$\frac{a-b}{2}=c$a−b2​=c	Simplify

 

We can repeat the proof above for the other two cases. The results prove the following theorem:

Practice questions

Question 3

Question 4