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Rule

Angles Outside the Circle Theorem

If a tangent and a secant, two tangents, or two secants intersect outside a circle, then the measure of the angle formed is one-half the difference of the measures of the intercepted arcs.

Based on the above diagrams, the following relations hold true.

m ∠ 1 m ∠ 2 m ∠ 3 = \frac{1}{2} (m B C - m A C) = \frac{1}{2} (m D E F - m D F) = \frac{1}{2} (m H I - m G J)

Below, a proof of the first equation is shown. The other two equations can be proven using similar reasoning.

Proof

Consider the first diagram and draw the auxiliary chord $\overline{B C}$ and point $P .$

By the Inscribed Angle Theorem, the measure of

∠ A B C

is half the measure of

A C .

Also, the Tangent and Intersected Chord Theorem tells that the measure of

∠ P C B

is half the measure of

B C .

m ∠ A B C m ∠ P C B = \frac{1}{2} m A C = \frac{1}{2} m B C

Next, the Triangle Exterior Angle Theorem says that

m ∠ P C B

equals the sum of the measures of

∠ A B C

and

∠ 1 .

Express the measure of

∠ 1

using this information.

m ∠ P C B = m ∠ A B C + m ∠ 1 ⇓ m ∠ 1 = m ∠ P C B - m ∠ A B C

Next, substitute the two angle measures found before into the right-hand side.

m ∠ 1 = m ∠ P C B - m ∠ A B C ⇓ m ∠ 1 = \frac{1}{2} m B C - \frac{1}{2} m A C

Finally, the proof can be completed by factoring

\frac{1}{2}

out.

$m ∠ 1 = \frac{1}{2} (m B C - m A C)$

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