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{{ printedBook.courseTrack.name }} {{ printedBook.name }} The measure of an inscribed angle is half that of its intercepted arc.

In the diagram above, the measure of $∠BAC$ is half the measure of arc $BC.$ This can be proven with the Isosceles Triangle Theorem.An inscribed angle can be created in three different ways, all depending on where the angle is located in relation to the center of the circle.

- The center is on a side of the inscribed angle.
- The center is inside the inscribed angle.
- The center is outside the inscribed angle.

Therefore, it must be proven that the theorem applies for all types of inscribed angles.

Consider an inscribed angle where the center of the circle lies on a side of the inscribe angle.

Then, there is a triangle created, $△MAB.$ Since two of the triangle's sides are the radius of the circle, it's an isosceles triangle.

In an isosceles triangle, the opposite angles to the congruent sides are congruent as well, by the Isosceles Triangle Theorem. Thus, the measure of $∠MBA$ is equal to $α.$

By the Interior Angles Theorem, the sum of the angles in $△MBA$ is $180_{∘}.$ Therefore, the last angle in the triangle, $∠AMB,$ can be expressed as: $180−α−α=180−2α.$

The angle $∠ABC$ is a straight angle and measures to $180_{∘}.$ Since it's the sum of $∠AMB$ and $∠CMB,$ the following equation can be written: $(180_{∘}−2α)+θ=180_{∘}.$ The expression is now simplified to prove the theorem.$(180_{∘}−2α)+θ=180_{∘}$

$θ=2α$

The next type is when the center of the circle is located inside the inscribed angle.

A diameter can be drawn from $A$ to the other side of the circle, dividing the inscribed angle into two.

Thus, two inscribed triangles are created with one of their sides through the center of the circle. It has already been proven that inscribed angles with one side through the circle's center satisfies the theorem. $θ_{1}=2α_{1},θ_{2}=2α_{2}$ Now, the sum of the measures of the smaller inscribed angles equals the larger one. The same applies for the smaller interpreted arcs. Hence, the relation between the angles can be expressed as: $θ_{1}+θ_{2}=2α_{1}+2α_{2}.$ Now by substituting the sum of the smaller angles with the larger ones, the theorem is given. $θ_{1}+θ_{2}=2(α_{1}+α_{2})⇔θ=2α$

The last type of inscribed angle is when the center of the circle lies outside the angle.

Similar to before, a diameter is drawn from $A$ through the midpoint.

The diameter creates a new inscribed angle with one side through the center of the circle.

It has previously been proven that inscribed angles with one side through the circle's center satisfies the theorem. $ω=2δ.$ Further, another inscribed angle is created as well. It's the sum of the two smaller inscribed angles.

Since one side goes through the center of the circle, the theorem applies. $θ+ω=2(α+ω)$ By substituting $δ$ with $2ω,$ it can be proven that the theorem applies for $α$ and $θ.$$θ+ω=2(α+δ)$

Substitute$ω=2δ$

$θ+2δ=2(α+δ)$

MultParMultiply parentheses

$θ+2δ=2α+2δ$

SubEqn$LHS−2δ=RHS−2δ$

$θ=2α$