Sign In
| 17 Theory slides |
| 11 Exercises - Grade E - A |
| Each lesson is meant to take 1-2 classroom sessions |
On the circle above, construct an angle with a vertex at the center of the circle. The angle being constructed should also cut off the same arc. In other words, construct the corresponding central angle.
Observe the measures of the angle and the arc intercepted by the angle. Start by moving B so that B, O, and A are collinear. Then, move it once again so that B, O, and C are collinear.
As can be seen, when B, O, and C are collinear, BC becomes the diameter of the circle, and the angle cuts off the semicircle. Furthermore, the measure of BC is twice the measure of an inscribed angle that intercepts it. This statement can be restated as a theorem.
The measure of an inscribed angle is half the measure of its intercepted arc.
In this figure, the measure of ∠BAC is half the measure of BC.
m∠BAC=21mBC
Let the measures of ∠BAD and ∠DAC be α and β, respectively.
Because the radii of a circle are all congruent, two isosceles triangles can be obtained by drawing OB and OC.
Therefore, by the Isosceles Triangle Theorem, the measures of ∠OBA and ∠OCA will also be α and β, respectively.
By the Triangle Exterior Angle Theorem, it is recognized that m∠BOD=2α and m∠COD=2β.
m∠BAC=21mBC
By applying similar logic as the procedure above, Case I and Case III can be proven.
In the diagram, the vertex of ∠UVT is on the circle O, and the sides of the angle are chords of the circle. Given the measure of UT, find the measure of the angle.
Write the answer without the degree symbol.
The Inscribed Angle Theorem can be used to find the measure of the angle.
The angle shown in the diagram fits the definition of an inscribed angle. For this reason, the measure of the angle can be found using the Inscribed Angle Theorem. The theorem states that the measure of an inscribed angle is half the measure of its intercepted arc.
Find the measure of the inscribed angle in the circle.
Similarly, given the measure of an inscribed angle, the measure of its corresponding central angle can be found using the Inscribed Angle Theorem. This can be done because the measure of the central angle is the same as the measure of the arc that the central angle cuts off.
In the circle, ∠KLM measures 67∘.
Find the measure of the corresponding central angle.
Start by drawing the corresponding central angle.
Recall that a central angle is an angle whose vertex lies at the center of the circle. Additionally, the inscribed angle and its corresponding central angle intercept the same arc KM for this example. Therefore, the corresponding central angle is ∠KOM.
Given the measure of an inscribed angle, find the measure of its corresponding central angle.
Up to now, the relationship between inscribed angles and their corresponding central angles has been discussed. Now the relationship between two inscribed angles that intercept the same arc will be investigated.
As can be observed, the angles are congruent, so long as they intercept the same arc.
If two inscribed angles of a circle intercept the same arc, then they are congruent.
By this theorem, ∠ADB and ∠ACB in the above diagram are congruent angles.
∠ADB≅∠ACB
Consider two inscribed angles ∠ADB and ∠ACB that intercept the same arc AB in a circle.
∠ADB≅∠ACB
Mark and Jordan have been asked to find the measure of ∠G.
Determine which angles intercept the same arc. Use the Inscribed Angles of a Circle Theorem to find m∠G.
The inscribed angles E and H intercept FG.
Inscribed angles, or the central angles, are not the only angles related to circles. In the next part, the angles constructed outside the circles will be examined. To construct an angle outside a circle, tangents can be used.
A line is a tangent to a circle if and only if the line is perpendicular to the endpoint of a radius on the circle's circumference.
Based on the diagram, the following relation holds true.
Line m is tangent to ⊙Q ⇔ m⊥QP
The theorem will be proven in two parts as it is a biconditional statement. Each will be proven by using an indirect proof.
Assume that line m is tangent to the circle centered at Q and not perpendicular to QP. By the Perpendicular Postulate, there is another segment from Q that is perpendicular to m. Let that segment be QT. The goal is to prove that QP must be that segment. The following diagram shows the mentioned characteristics.
Line m is tangent to ⊙Q ⇒ m⊥QP
For the second part, it will be assumed that m is perpendicular to the radius QP at P, and that line m is not tangent to ⊙Q. In this case, line m intersects ⊙Q at a second point R.
m⊥QP ⇒ line m is tangent to ⊙Q
Having proven both parts, the proof of the biconditional statement of the theorem is now complete.
Line m is tangent to ⊙Q ⇔ m⊥QP
In the diagram, ℓ is tangent to the circle at the point A, and DA is a diameter.
A tangent is perpendicular to the radius of a circle through the point of tangency.
It has been given that m∠ABC=60∘. By the Tangent to Circle Theorem, ℓ is perpendicular to DA. In other words, m∠BAD=90∘.
A circumscribed angle is supplementary to the central angle it cuts off.
The measure of a circumscribed angle is equal to 180∘ minus the measure of the central angle that intercepts the same arc.
Considering the above diagram, the following relation holds true.
m∠ADB=180∘−m∠ACB
By definition, a circumscribed angle is an angle whose sides are tangents to a circle. Since ∠ADB is a circumscribed angle, DA and DB are tangents to ⊙C at points A and B, respectively. By the Tangent to Circle Theorem, CA is perpendicular to DA and CB is perpendicular to DB.
m∠DAC=90∘, m∠CBD=90∘
Find the measure of the central angle.
The following example involving circumscribed angles and inscribed angles could require the use of the previously learned theorems.
Two tangents from P to ⊙M are drawn. The measure of ∠KPL is 40∘.
Find the measure of the inscribed angle that intercepts the same arc as ∠KPL?
Use the Circumscribed Angle Theorem and Inscribed Angle Theorem.
This lesson defined three angles related to circles as well as the relationships between these angles. The diagram below shows the definitions and the main theorems of this lesson.
In the following diagram, BC is tangent to ⊙M.
We know that BC is tangent to ⊙ M. According to the Tangent to Circle Theorem, it is then perpendicular to the circle's radius. Therefore, ∠ FAC is a right angle. Since ∠ EAC is 56^(∘), the measure of ∠ FAE is the difference between the right angle and 56^(∘). m∠ FAE = 90^(∘) - 56^(∘) = 34^(∘) Let's add this to the diagram.
If we draw EM, we get the isosceles triangle EMA. We know that it is isosceles because two of its sides are the radius of the circle which means they are congruent. According to the Base Angles Theorem, the base angles of an isosceles triangle are congruent.
Using the Interior Angles Theorem, we can calculate the triangle's vertex angle. m∠ M = 180^(∘) -2(34^(∘)) =112^(∘) Notice that ∠ EDA and ∠ EMA are the inscribed angle and central angle to the intercepted arc EA. The central angle and its intercepted arc always have the same measure. Let's add this to the diagram.
Now we can use the Inscribed Angle Theorem to find m ∠ EDA.
In Part A, we were given a numerical value of ∠ EAC. Now we are working with a general angle, x. Similar to Part A, we can calculate ∠ FAE by subtracting x from 90^(∘).
m∠ FAE = 90^(∘) - x
Let's add this to the diagram. We will also add y to the diagram.
Like in Part A, we can form an isosceles triangle by drawing EM. The second base angle will be congruent to ∠ FAE.
When we have expressions for the triangle's base angles, we can determine the vertex angle. m∠ M = 180^(∘) -2(90^(∘)-x) =2x Again, since the central angle has the same measure as its intercepted arc, we can use the Inscribed Angles Theorem to write a relationship between x and y.
In the following diagram, AB and BC are tangents to ⊙M.
Let's draw radii from M to each point of tangency and label them D and E. According to the Tangent to Circle Theorem, MD and DE must intersect AB and BC at right angles.
Notice that BDME is a quadrilateral which means it has an angle sum of 360^(∘). We know the measure of two of these angles. We can use them to write an expression for m∠ DME.
Let's add this equation to the diagram.
Notice that the reflexive angle at M is the difference between 360^(∘) and 180^(∘)-x. 360^(∘)-(180^(∘)-x)=180^(∘)+x Let's add this information to the diagram. Additionally, take notice that this is the central angle to the intercepted arc DE. Therefore, these angles are congruent.
Now we can use the Inscribed Angle Theorem to find x.