Exploring Circle Theorems: Outside Angle Theorem and Inscribed Angle Theorem

The theorem will be proven in two parts as it is a biconditional statement. Each will be proven by using an indirect proof.

Part I: If a Line Is a Tangent to a Circle, Then It Is Perpendicular to the Endpoint of a Radius on the Circle’s Circumference

Assume that line $m$ is tangent to the circle centered at $Q$ and not perpendicular to $\overline{Q P} .$ By the Perpendicular Postulate, there is another segment from $Q$ that is perpendicular to $m .$ Let that segment be $\overline{Q T} .$ The goal is to prove that $\overline{Q P}$ must be that segment. The following diagram shows the mentioned characteristics.

Since

\overline{Q T}

is said to be perpendicular to

m,

△ Q T P

is a right triangle. In this case,

\overline{Q P},

whose length is

r

units, is the hypotenuse and, therefore, the longest side. As a result,

Q T

is less than

Q P .

Q T < Q P

There is a part of

\overline{Q T}

that lies outside of the circle. Let

b

be the length of that part. The other part of

\overline{Q T}

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,

Q T

is equal to the sum of

r

and

b .

This expression, together with

Q P = r,

can be substituted into the aforementioned inequality.

Q T < Q P

SubstituteExpressions

Substitute expressions

r + b < r

SubIneq

$LHS - r < RHS - r$

b < 0

Since a length cannot be less than

0,

the assumption that

\overline{Q P}

is not perpendicular to the line

m

must be false. Therefore,

\overline{Q P}

is perpendicular to the tangent

m

at its endpoint on the circle. This concludes the first part of the proof.

Line $m$ is tangent to $⊙ Q$ $\Rightarrow$ $m ⊥ \overline{Q P}$

Part II: If a Line Is Perpendicular to the Endpoint of a Radius on a Circle’s Circumference, Then It Is a Tangent to the Circle

For the second part, it will be assumed that $m$ is perpendicular to the radius $\overline{Q P}$ at $P,$ and that line $m$ is not tangent to $⊙ Q .$ In this case, line $m$ intersects $⊙ Q$ at a second point $R .$

Since

m

is said to be perpendicular to

\overline{Q P}

P,

△ Q P R

is a right triangle and

\overline{Q R}

is the hypotenuse. Therefore,

Q P

is less than

Q R .

Q P < Q R

However, it can also be seen that both

\overline{Q P}

and

\overline{Q R}

are radii of

⊙ Q .

Therefore, they have the same length.

Q P = Q R

These two statements are contradictory to each other. Therefore, both cannot be true. The contradiction came from supposing that line

m

was not tangent to

⊙ Q .

Consequently,

m

must be a tangent line to the circle. This completes the second part of the proof.

$m ⊥ \overline{Q P}$ $\Rightarrow$ 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 $⊙ Q$ $\Leftrightarrow$ $m ⊥ \overline{Q P}$

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Catch-Up and Review

Investigating Inscribed Angles of a Circle

Inscribed and Central Angles of a Circle

Inscribed Angle of a Circle

Inscribed Angle Theorem

Proof

Finding the Inscribed Angle of a Circle

Hint

Solution

Practice Finding Inscribed Angles

Finding the Central Angle of a Circle

Hint

Solution

Practice Finding Central Angles

Inscribed Angles of a Circle Theorem

Proof

Using the Inscribed Angles of a Circle Theorem

Hint

Solution

Constructing Angles Outside of a Circle

Tangent

Tangent to a Circle Theorem

Proof

Part I: If a Line Is a Tangent to a Circle, Then It Is Perpendicular to the Endpoint of a Radius on the Circle’s Circumference

Part II: If a Line Is Perpendicular to the Endpoint of a Radius on a Circle’s Circumference, Then It Is a Tangent to the Circle

Using the Tangent to Circle Theorem

Hint

Solution

Circumscribed Angle Theorem

Proof

Practice Using the Circumscribed Angle Theorem

Using Circle Theorems to Solve Problems

Hint

Solution

Summarizing Angle Theorems Related to Circles

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