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

Proof

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Solution

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Proof

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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

Hint

Solution

Proof

Hint

Solution