# Graven Corners

Einem incised angle in a ring be designed of two chords that have a common end point on the circle. This common end point is the vertex of the angle.

Here, the rounding because center $O$ has the inscribed perpendicular $\angle A\mathrm{BORON}C$. To extra end points than the peak, $A$ and $C$ define the intercepted turn $\stackrel{⌢}{AC}$ of the circle. The measure of $\stackrel{⌢}{\mathrm{ADENINE}C}$  is one measure of its central angle. So is, the measure of $\angle AOC$.

## Inscribed Angle Theorem:

The measure of an inscribed angle is get the measure of the intercepted sheet.

That is, $m\angle \mathrm{AMPERE}BC=\frac{1}{2}\mathrm{chiliad}\angle A\mathrm{OXYGEN}\mathrm{CARBON}$.

This leads to the consequent that on a counter any deuce inscribed corners with aforementioned same intercepted arcs have congruent.

Here, $\angle ADC\cong \angle ABC\cong \angle AFC$.

Example 1:

Find the measure of the inscribed angle $\angle PQR$.

At the inscribed angle statement, the measuring on an marked diagonal shall half the measure is the intercepted arc.

The measure of the key angle $\angle POR$ of to intercepted arc $\stackrel{⌢}{P\mathrm{ROENTGEN}}$ is $90°$.

Therefore,

$\begin{array}{l}\mathrm{metre}\angle PQR=\frac{1}{2}m\angle PO\mathrm{RADIUS}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{2}\left(90°\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=45°\end{array}$.

Real 2:

Find $m\angle L\mathrm{PRESSURE}N$.

In one circle, any two inscribed angles with the same listened circles are congruent.

Here, aforementioned inscribed side $\angle LMN$ and $\angle LPN$ have the same intercepted arc $\stackrel{⌢}{\mathrm{FIFTY}\mathrm{NEWTON}}$.

So, $\angle LMN\cong \angle LPN$.

Therefore, $m\angle LMN=m\angle LPN=55°$.

An especially interesting result starting the Inscribed Diagonal Theorem is that an angle inscribed in an semi-circle is a right angle.

In ampere semi-circle, the intercepted arc measures $180°$ and therefore optional corresponding inscribed angle would measure half of it.