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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.

Recall that the measure of an arc is the measure of the central angle that includes the arc. Now, move the points and make conjectures that explain the relationship between the measures of angles and arcs.

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 $BC.$

$m∠BAC=21 mBC$

An inscribed angle can be created in three different ways depending on where the angle is located in relation to the center of the circle.

Consider Case II. A diameter can be drawn from $A$ through $O,$ dividing the inscribed angle into two.

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 $m∠COD=2β.$

The measure of an intercepted arc is the same as the measure of its central angle. Therefore, the measures of $BD$ and $CD$ are $2α$ and $2β,$ respectively.$mBD=2αandmCD=2β $

From here, using the Angle Addition Postulate and the Arc Addition Postulate, $m∠BAC$ and $mBC$ can be written in terms of $α$ and $β.$
$m∠BACmBCmBC =α+β=2α+2β=2(α+β) $

Consequentially, the measure of $∠BAC$ is half the measure of $mBC.$ The proof of Case II has been completed. $m∠BAC=21 mBC$

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.

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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.

The arc intercepted by $∠UVT$ measures $40_{∘}.$ Therefore, $∠UVT$ measures $20_{∘}.$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.

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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 $∠KOM.$

The arc $KM$ is included in $∠KOM.$ Therefore, the measure of $∠KOM$ is equal to the measure of $KM.$ Using the Inscribed Angle Theorem, the measure of $KM$ can be found. Therefore, $∠KOM$ also measures $134_{∘}.$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.

By the Inscribed Angle Theorem, the measures of $∠ADB$ and $∠ACB$ are half the measure of $AB.$$m∠ADB=21 mABm∠ACB=21 mAB $

Therefore, by the Transitive Property of Equality, it can be said that these two angles have the same measure.
$m∠ADB=m∠ACB $

Consequently, by the definition of congruent angles, $∠ADB$ and $∠ACB$ are congruent. $∠ADB≅∠ACB$

Mark and Jordan have been asked to find the measure of $∠G.$

Mark claims that the measure of $∠G$ is the same as the measure of $∠E.$ Jordan, however, thinks that its measure is half the measure of $∠E.$ Who is correct?{"type":"choice","form":{"alts":["Mark","Neither","Jordan","Both"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}

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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 $FG.$

By the Inscribed Angles of a Circle Theorem, $∠E$ is congruent to $∠H.$$∠E≅∠H⇕m∠E=m∠H=48_{∘} $

Accordingly, half the measure of $∠E$ can be found as $248_{∘} =24_{∘}.$ Additionally, $∠F$ and $∠G$ intercept the same arc, $EH.$
Again by the Inscribed Angles of a Circle Theorem, $∠F$ is congruent to $∠G.$ Therefore, $m∠G=25_{∘}.$ $∠F≅∠G⇕m∠F=25_{∘}=m∠G $

Since $m∠G=25_{∘},$ neither Mark's claim nor Jordan's claim is correct. 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.

$QT<QP $

There is a part of $QT $ that lies outside of the circle. Let $b$ be the length of that part. The other part of $QT $ is the radius of $⊙Q,$ because it is a segment from the center of the circle to a point on the circle.
By the Segment Addition Postulate, $QT$ is equal to the sum of $r$ and $b.$ This expression, together with $QP=r,$ can be substituted into the aforementioned inequality.
Since a length cannot be less than $0,$ the assumption that $QP $ is 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 $R.$

$QP<QR $

However, it can also be seen that both $QP $ and $QR $ are radii of $⊙Q.$ Therefore, they have the same length.
$QP=QR $

These two statements are contradictory to each other. Therefore, both $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.

If the measure of $∠ABC$ is $60_{∘}.$ Find the measure of $DC.${"type":"choice","form":{"alts":["<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\">3<\/span><span class=\"mord\"><span class=\"mord\">0<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\">6<\/span><span class=\"mord\"><span class=\"mord\">0<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\">9<\/span><span class=\"mord\"><span class=\"mord\">0<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>","<span class=\"katex\"><span class=\"katex-html\" aria-hidden=\"true\"><span class=\"base\"><span class=\"strut\" style=\"height:0.674115em;vertical-align:0em;\"><\/span><span class=\"mord\">1<\/span><span class=\"mord\">2<\/span><span class=\"mord\"><span class=\"mord\">0<\/span><span class=\"msupsub\"><span class=\"vlist-t\"><span class=\"vlist-r\"><span class=\"vlist\" style=\"height:0.674115em;\"><span style=\"top:-3.063em;margin-right:0.05em;\"><span class=\"pstrut\" style=\"height:2.7em;\"><\/span><span class=\"sizing reset-size6 size3 mtight\"><span class=\"mbin mtight\">\u2218<\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span>"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":3}

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, $m∠BAD=90_{∘}.$

Since the sum of the measures of the interior angles of a triangle is $180_{∘},$ $m∠D=30_{∘}.$$90_{∘}+60_{∘}+m∠ADBm∠ADB =180_{∘}=30_{∘} $

Notice that $∠ACD$ is an inscribed angle on the diameter $AD$ of $⊙E.$ Therefore, $∠ACD$ measures $90_{∘}.$
Once again, using the Triangle Sum Theorem, the measure of $∠DAC$ can be found to be $60_{∘}.$ $90_{∘}+30_{∘}+m∠DACm∠DAC =180_{∘}=60_{∘} $

Since $DC$ is associated with $∠DAC$ and $m∠DAC=60_{∘},$ the measure of the arc can be found using the Inscribed Angle Theorem.
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 $DB.$

Notice that $ADBC$ is a quadrilateral and two of its angles are right angles. Recall that the sum of all of the angles in a quadrilateral is $360_{∘}.$ Substituting the known angle measures and solving for $m∠ADB$ will give the desired equation.$m∠ADB+m∠DAC+m∠ACB+m∠CBD=360_{∘}$

SubstituteII

$m∠DAC=90_{∘}$, $m∠CBD=90_{∘}$

$m∠ADB+90_{∘}+m∠ACB+90_{∘}=360_{∘}$

$m∠ADB=180_{∘}−m∠ACB$

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 $∠KPL?$

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Use the Circumscribed Angle Theorem and Inscribed Angle Theorem.

The inscribed angle that intercepts $KL$ is $∠KNL.$ By the Inscribed Angle Theorem, $m∠KNL$ is half $mKL.$

$m∠KNL=21 mKL $

Recall that the measure of an arc is the same as the measure of its corresponding central angle. Therefore, $m∠KML$ is equal to $mKL.$
Substitute $m∠KML$ into the above equation.
$m∠KNLm∠KNL =21 mKL=21 m∠KML $

Now, by the Circumscribed Angle Theorem, $m∠KML$ is $180_{∘}$ minus $m∠KPL.$ $m∠KNL=21 (180_{∘}−m∠KPL) $

It is given that $m∠KPL=40_{∘}.$ Substituting that measure into the equation will give the measure of the inscribed angle.
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.