Angles Inside the Circle Theorem

Let $E$ be the point of intersection of chords $\overline{A C}$ and $\overline{B D} .$ Start by drawing an auxiliary chord $\overline{B C} .$

Notice that

∠ 1

is an exterior angle of

△ B C E .

Therefore, by the Triangle Exterior Angle Theorem, its measure equals the sum of the measures of the two non-adjacent interior angles.

m ∠ 1 = m ∠ A C B + m ∠ C B D

By the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc.

{m ∠ A C B = \frac{1}{2} m A B m ∠ C B D = \frac{1}{2} m C D (I) (II)

Finally, substituting these two equations into the equation given by the Triangle Exterior Angle Theorem, the first required equation is obtained.

m ∠ 1 = \frac{1}{2} m A B + \frac{1}{2} m C D ⇓ m ∠ 1 = \frac{1}{2} (m A B + m C D)

To obtain the second equation, draw the auxiliary chord $\overline{A B} .$

As before,

∠ 2

is an exterior angle of

△ A B E .

Therefore, its measure is equal to the sum of the measures of the two non-adjacent interior angles.

m ∠ 2 = m ∠ A B D + m ∠ C A B

Once more, the Inscribed Angle Theorem can be applied to rewrite the two angle measures on the right-hand side in terms of the corresponding intercepted arcs.

{m ∠ A B D = \frac{1}{2} m A D m ∠ C A B = \frac{1}{2} m B C (I) (II)

Finally, substitute the last two equations into the one relating

m ∠ 2

and the measure of the inscribed angles.

m ∠ 2 = m ∠ A B D + m ∠ C A B ⇓ m ∠ 2 = \frac{1}{2} (m A D + m B C)

{{ article.displayTitle }}

Angles Inside the Circle Theorem

Proof

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount }}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount }}
	{{ 'ml-lesson-time-estimation' \| message }}