We will start this lesson with a list of important theorems and what they say. Unlike in most lessons, we will not fully prove all of them here, as there are a great many of them, so we encourage you to try to prove the remaining theorems yourself and to look up proofs online if you can't. In addition, because there are so many theorems and we want this page to be organized, the lesson will be presented slightly differently. Even though all of these theorems aren't in sample problem format, we encourage you to attempt to prove them yourself before reading the solution, and we encourage you to try to look at the methods used to prove these theorems as general strategies that can be used in many problems involving circles (ex., extending radii out to tangents is always a good idea). Finally, the names of these theorems are not that important to memorize, so long as you are familiar with what the theorem does.
Inscribed Angle Theorem
![Picture](/uploads/1/1/2/8/112821707/published/cmathg5.png?1511376426)
This theorem tells us that an angle inscribed in a circle (its vertex and sides intersect the circle) has an angle measure half that of the arc it intercepts (the arc formed by the intersection points of the circle and the side of the angle). As seen in the diagram to the left, we would have that \(m\measuredangle HIJ=\frac{m\measuredangle JGH}{2}\) (the measure of the central angle, the one formed by connecting the arc's endpoints to the center of the circle. is equal to the measure of the arc).Note that this theorem also implies any two angles intersecting the same arc are congruent.
We can prove this theorem by drawing the diameter of the circle through the vertex of the inscribed angle and doing casework on if the end points of the arc are on opposite sides of the diameter, one is on the diameter and the other isn't, or if both are on the same side of the diameter. The first to proofs are fairly trivial and for the third proof we can use our second proof to make it fairly easy as well. We encourage you to work out the details yourself.
We can prove this theorem by drawing the diameter of the circle through the vertex of the inscribed angle and doing casework on if the end points of the arc are on opposite sides of the diameter, one is on the diameter and the other isn't, or if both are on the same side of the diameter. The first to proofs are fairly trivial and for the third proof we can use our second proof to make it fairly easy as well. We encourage you to work out the details yourself.
A bunch of theorems about secants, chords, tangents, and angle measures with insignificant names
![Picture](/uploads/1/1/2/8/112821707/published/cmathg7_3.png?1511392033)
Some theorem states that the measure of the angle a chord forms by intersecting a tangent at the point of tangency is half the angle of the arc in intercepts.In the diagram to the left, this would mean that \(m\measuredangle CBD=\frac{m\measuredangle CAB}{2}\).
To prove this theorem, we will start by noting that CB is clearly the diameter of the circle because it is perpendicular to the tangent. Thus, the measure of arc BE is equal to \(180^{\circ} -m\measuredangle CAE\). Note that, because triangle BAE is isosceles, \(m\measuredangle ABE=\frac{m\measuredangle CAE}{2}\). Thus, \(m\measuredangle EBD=90^{\circ} -\frac{m\measuredangle CAE}{2}\). However, this means that \(m\measuredangle EBD = \frac{m\measuredangle BAE}{2}\). Thus, because BE was an arbitrary chord, we have proved the desired theorem.
To prove this theorem, we will start by noting that CB is clearly the diameter of the circle because it is perpendicular to the tangent. Thus, the measure of arc BE is equal to \(180^{\circ} -m\measuredangle CAE\). Note that, because triangle BAE is isosceles, \(m\measuredangle ABE=\frac{m\measuredangle CAE}{2}\). Thus, \(m\measuredangle EBD=90^{\circ} -\frac{m\measuredangle CAE}{2}\). However, this means that \(m\measuredangle EBD = \frac{m\measuredangle BAE}{2}\). Thus, because BE was an arbitrary chord, we have proved the desired theorem.
![Picture](/uploads/1/1/2/8/112821707/published/cmathg8.png?1511391976)
Another theorem states that the measure of the angle that two secants form where they intersect is equal to half the difference in angle measures of the two arcs in the circle that the secants intersect. For example, in the diagram to the left, \(m\measuredangle FDE=\frac{m\measuredangle GAH-m\measuredangle FAE}{2}\).
We will prove this theorem with the configuration where the center of the circle is in between the secants and other configurations can be handled similarly (although we encourage you to prove them on your own). First of all, it is apparent that quadrilateral FAED has two angles that we want to know about, so we want to find all of the angles in the quadrilateral in terms of angles we want to end up using. Looking at angles AFD and DEA we see that we can easily relate them to angles GAE and HAF using the fact that triangles GAE and HAF are isosceles. Next, we note that even though GAE and HAF aren't angles we want to end up in our answer, the fact that \(m\measuredangle FAE + m\measuredangle FAH + m\measuredangle GAE+m\measuredangle HAG=360^{\circ}\) could help us get our answer in terms of the angles we want. Note that \(m\measuredangle AED=180^{\circ}-m\measuredangle AEG=90^{\circ}+\frac{m\measuredangle GAE}{2}\). Similarly, \(m\measuredangle AFD=90^{\circ}+\frac{m\measuredangle FAD}{2}\). Thus, \(m\measuredangle FDE = 180^{\circ}-m\measuredangle FAE -\frac{m\measuredangle FAH}{2}-\frac{m\measuredangle FAH}{2}=\frac{360^{\circ}- m\measuredangle GAE- m\measuredangle FAH}{2}-m\measuredangle FAE =\frac{m\measuredangle GAH+m\measuredangle FAE}{2}-m\measuredangle FAE=\frac{m\measuredangle GAH- m\measuredangle FAE}{2}\).
The same theorem holds true for the a tangent and a secant that intersect outside of a circle as well as a two tangents that intersect outside of a circle. The theorems say essentially the exact same thing and can be proved in a very similar way.
We will prove this theorem with the configuration where the center of the circle is in between the secants and other configurations can be handled similarly (although we encourage you to prove them on your own). First of all, it is apparent that quadrilateral FAED has two angles that we want to know about, so we want to find all of the angles in the quadrilateral in terms of angles we want to end up using. Looking at angles AFD and DEA we see that we can easily relate them to angles GAE and HAF using the fact that triangles GAE and HAF are isosceles. Next, we note that even though GAE and HAF aren't angles we want to end up in our answer, the fact that \(m\measuredangle FAE + m\measuredangle FAH + m\measuredangle GAE+m\measuredangle HAG=360^{\circ}\) could help us get our answer in terms of the angles we want. Note that \(m\measuredangle AED=180^{\circ}-m\measuredangle AEG=90^{\circ}+\frac{m\measuredangle GAE}{2}\). Similarly, \(m\measuredangle AFD=90^{\circ}+\frac{m\measuredangle FAD}{2}\). Thus, \(m\measuredangle FDE = 180^{\circ}-m\measuredangle FAE -\frac{m\measuredangle FAH}{2}-\frac{m\measuredangle FAH}{2}=\frac{360^{\circ}- m\measuredangle GAE- m\measuredangle FAH}{2}-m\measuredangle FAE =\frac{m\measuredangle GAH+m\measuredangle FAE}{2}-m\measuredangle FAE=\frac{m\measuredangle GAH- m\measuredangle FAE}{2}\).
The same theorem holds true for the a tangent and a secant that intersect outside of a circle as well as a two tangents that intersect outside of a circle. The theorems say essentially the exact same thing and can be proved in a very similar way.
![Picture](/uploads/1/1/2/8/112821707/published/cmathg9.png?1511395747)
A theorem states that when two chords intersect each other inside of a circle, the measure of the angle of their intersection is equal to the average of the measure of the arc that the angles intersects. In the diagram to the left, this would mean that \(m\measuredangle CFE=m\measuredangle DFB=\frac{m\measuredangle DOB + m\measuredangle COE}{2}\).
You will prove this theorem in the exercises.
You will prove this theorem in the exercises.
Power of a Point
This theorem has a different definition than we will cover here, but as far as your use of it will go, this definition will serve to be all you need. The theorem has several parts.
![Picture](/uploads/1/1/2/8/112821707/published/cmathg4.png?1511373831)
In this case, the theorem states that if we draw any chord through a point in a circle, The product of the distance from that point to the two intersections of that chord with the circle will stay constant. In the diagram, the theorem tells us that \(FB\cdot FC=FD\cdot FE\).
To prove this theorem, we can just draw two arbitrary chords through an arbitrary point and prove that these for these to chord the theorem holds. This will prove the whole theorem because this means that for any point, if we have one chord, the desired product will be equal to the product for that chord in any other chord that we draw. To prove that this holds, we will use the picture to the left as our diagram. We can see that triangles EBF and CDF are similar using the fact that angle BEF and DCF both intercept the same arc-and are thus congruent by our previous our previous theorem and angles EBF and CDF are similarly congruent. Thus, we get \(\frac{FE}{FB}=\frac{FC}{FD}\implies FB\cdot FC=FD\cdot FE\).
To prove this theorem, we can just draw two arbitrary chords through an arbitrary point and prove that these for these to chord the theorem holds. This will prove the whole theorem because this means that for any point, if we have one chord, the desired product will be equal to the product for that chord in any other chord that we draw. To prove that this holds, we will use the picture to the left as our diagram. We can see that triangles EBF and CDF are similar using the fact that angle BEF and DCF both intercept the same arc-and are thus congruent by our previous our previous theorem and angles EBF and CDF are similarly congruent. Thus, we get \(\frac{FE}{FB}=\frac{FC}{FD}\implies FB\cdot FC=FD\cdot FE\).
![Picture](/uploads/1/1/2/8/112821707/published/cmathg6_1.png?1511398074)
In this case, this theorem would state that For any two secant to a circle that intersect at a point outside of the circle, the product the the distances from the point outside the circle to the two intersection points of the secant going through hath point and the circle is equal. In the diagram to the left, this would mean that \(CF\cdot CE = CG\cdot CD\).
You Will prove this in the excercises
You Will prove this in the excercises
![Picture](/uploads/1/1/2/8/112821707/published/cmathg10_1.png?1511397071)
Next, in this case, the theorem states that the product of the distances from the intersect of a tangent and a secant to a circle to the two intersections of the tangent with the circle is equal to the distance from the same intersection point to the point of tangency squared. In the diagram to the left, this theorem would state that \(DB^2=DC\cdot DE\).
To prove this case, we will introduce a concept called the power of a point, the very concepts this theorem is named after. The power of point P with respect to circle O is the distance from P to O squared minus the length of the radius of circle O squared. Note that this quantity can be negative when P is inside circle O. For this proof, we will show that both of our desired expressions are equal to the power of point D with respect to circle A. First of all, by the Pythagorean Theorem, we have \(r^2+DB^2=DA^2\implies DB^2=DA^2-r^2\). Thus, \(DB^2\) is equal to the power of point D with respect to circle A. Next, note that \(DC\cdot DE=(DA-r)(DA+r)=DA^2-r^2\). Thus, \(DC\cdot DE\) is equal to the power of point D with respect to circle A and we are done. Note that this implies that if two tangents intersect at a point outside of a circle, then the distance from that point to the points of tang ency is the same for each tangent.
To prove this case, we will introduce a concept called the power of a point, the very concepts this theorem is named after. The power of point P with respect to circle O is the distance from P to O squared minus the length of the radius of circle O squared. Note that this quantity can be negative when P is inside circle O. For this proof, we will show that both of our desired expressions are equal to the power of point D with respect to circle A. First of all, by the Pythagorean Theorem, we have \(r^2+DB^2=DA^2\implies DB^2=DA^2-r^2\). Thus, \(DB^2\) is equal to the power of point D with respect to circle A. Next, note that \(DC\cdot DE=(DA-r)(DA+r)=DA^2-r^2\). Thus, \(DC\cdot DE\) is equal to the power of point D with respect to circle A and we are done. Note that this implies that if two tangents intersect at a point outside of a circle, then the distance from that point to the points of tang ency is the same for each tangent.
Exercises
- Prove Theorem 3 (its diagram is labeled Thm. 3 in red).
- Prove the second case of the Power of a Point theorem.
Solution 1
Solution 2
We will simply note that triangle DFC is similar to triangle EGC because angles CEG and CDF because they intersect the same arc. Thus, we get \(\frac{CD}{CF}=\frac{CE}{CG}\implies CF\cdot CE = CG\cdot CD\).
OR
Similarly to how we proved the last case of the Power of a Point theorem, both \(CF\cdot CE\) and \(CG\cdot CD\) are equal to the power of point C with respect to circle O.
OR
Similarly to how we proved the last case of the Power of a Point theorem, both \(CF\cdot CE\) and \(CG\cdot CD\) are equal to the power of point C with respect to circle O.