Let's begin by writing what Theorem $10.9$ states.

Theorem $10.9$
If a quadrilateral is inscribed in a circle, then its opposite angles are supplementary.

Let's consider a circle centered at point $O$ and a quadrilateral $A B C D$ inscribed in it.

By the Arc Addition Postulate, the measures of

D C B

and

D A B

are

36 0^{\circ}

because they cover the entire circle.

m D C B + m D A B = 36 0^{\circ}

Also, by the Inscribed Angle Theorem the measure of an inscribed angle is half the measure of its intercepted arc.

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

Next, let's add these two equations.

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

FactorOut

Factor out $\frac{1}{2}$

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

▼

Simplify right-hand side

Substitute

$m D C B + m D A B = 36 0^{\circ}$

m ∠ A + m ∠ C = \frac{1}{2} (36 0^{\circ})

Multiply

m ∠ A + m ∠ C = 18 0^{\circ}

The last equation implies that

∠ A

and

∠ C

are supplementary. To show that

∠ B

and

∠ D

are also supplementary, we use the fact that the sum of the interior angles of a quadrilateral is

36 0^{\circ} .

m ∠ A + m ∠ B + m ∠ C + m ∠ D = 36 0^{\circ}

CommutativePropAdd

Commutative Property of Addition

m ∠ A + m ∠ C + m ∠ B + m ∠ D = 36 0^{\circ}

▼

Simplify right-hand side

Substitute

$m ∠ A + m ∠ C = 18 0^{\circ}$

18 0^{\circ} + m ∠ B + m ∠ D = 36 0^{\circ}

SubEqn

$LHS - 18 0^{\circ} = RHS - 18 0^{\circ}$

m ∠ B + m ∠ D = 18 0^{\circ}

Paragraph Proof

Given : Prove : Quadrilateral A B C D is inscribed in ⊙ O ∠ A and ∠ C are supplementary, ∠ B and ∠ D are supplementary

Proof: By the Arc Addition Postulate

m D C B + m D A B = 36 0^{\circ} .

Also, by the Inscribed Angle Theorem, we get

m ∠ A = \frac{1}{2} m D C B

and

m ∠ C = \frac{1}{2} m D A B .

Adding these two equation gives

m ∠ A + m ∠ C =

\frac{1}{2} (m D C B + m D A B) .

By substitution we have $m ∠ A + m ∠ C = \frac{1}{2} (36 0^{\circ}),$ or $18 0^{\circ} .$ This means that $∠ A$ and $∠ C$ are supplementary. Because the sum of the measures of the interior angles of a quadrilateral is $36 0^{\circ},$ we get $m ∠ A + m ∠ C +$ $m ∠ B +$ $m ∠ D = 36 0^{\circ} .$

Consequently $18 0^{\circ} + m ∠ B + m ∠ D = 36 0^{\circ},$ which implies that $m ∠ B + m ∠ D = 18 0^{\circ},$ making these angles supplementary as well.

Exercise