Inscribed angle is an angle formed by two line segments joining at a point. $\angle$ABC is an inscribed angle in the following figure.

AB and AC are two sides of the angle and point B where two side join is called vertex of the angle.
Angle

Inscribed angle of a circle is an angle formed by two chords which have one common endpoint. This common endpoint lies on the circumference of the circle. Other two endpoints of the chords intersect at the boundary of the circle.
The part of the circle between these two endpoints (together with the endpoints) is called intercepted arc. In the following figure, $\angle$PQR is an inscribed angle of a circle. Point Q is referred as vertex. Arc PR is intercepted arc and is denoted by $\widehat{PR}$.
Inscribed Angle of Circle
Following two angles are not inscribed angles of circle, since they are not satisfying the condition that one common endpoint must lie on the circumference of the circle.
Non Inscribed Angle
Inscribed angle formula enables us to evaluate the measure of intercepted arc. Measure of intercepted arc is equal to the angle subtended by intercepted arc at center of the circle. This angle is called central angle.

Inscribed angle formula states that the measure of inscribed angle is exactly equal to half of measure of intercepted arc. Inscribed angle formula can be given as follows:
Formula for Inscribed Angle
Angle inscribed in a semicircle is always a right angle.

The proof is given below:

Let us assume a semicircle with diameter AB.
Inscribed Angle of Semicircle
Given: ∠ACB is inscribed in semicircle.

To Prove: ∠ACB = 90°.

Construction: Join C and O.

Proof: Let ∠AOC = p and ∠BOC = q
∠p + ∠q = 180° (Linear pair angles)
AO = BO = CO (Radii)
Angles opposite to equal sides are equal.
∴ ∠CAO = ∠ACO = x(say)
 and ∠CBO = ∠BCO = y(say)

In ΔAOC:
∠x + ∠x + ∠p = 180° (Sum of the angles of a triangle is 180°)
2∠x + ∠p = 180°.......1
In ΔBOC:
∠y + ∠y + ∠q = 180° (Sum of the angles of a triangle is 180°)
2∠y + ∠q = 180°.......2
Adding equations 1 and 2, we get
2(∠x + ∠y) + ∠p + ∠q = 360°
2(∠x + ∠y) + 180° = 360°
2(∠x + ∠y) = 180°
∠x + ∠y = 90°
∠ACB = 90°
Hence, angle inscribed in a semicircle is always a right angle.
Inscribed angle theorem states, "Angle subtended by an arc at the center of a circle is double the angle subtended by that arc at any point on the circumference of the circle".

The proof of the theorem is as follows:
Inscribed Angle Theorem
Given: There are 3 possible cases.
Case 1: ∠AOB < 180°
Case 2: ∠AOB = 180°
Case 3: ∠AOB > 180°
In each case, Arc AEB subtends ∠AOB at the center and ∠ACB at a point C on the circle.

To Prove: ∠AOB = 2∠ACB

Construction: Join C to O and extend it to D.

Proof: In ΔAOC:
OA = OC (radii)
Since, Angles opposite to equal sides are equal.
∴ ∠OAC = ∠OCA = x(say)
In ΔBOC:
OB = OC
Since, Angles opposite to equal sides are equal.
∴ ∠OBC = ∠OCB = y(say)
∠AOD = ∠x + ∠x = 2∠x (Exterior angle of a triangle is equal to the sum of opposite interior angles)
∠BOD = ∠y + ∠y = 2∠y (Exterior angle of a triangle is equal to the sum of opposite interior angles)
∠AOB = ∠AOD + ∠BOD
∠AOB = 2∠x + 2∠y
∠AOB = 2(∠x + ∠y)
∠AOB = 2∠ACB
Few problems based on inscribed angle are as follows:

Solved Examples

Question 1: Find the value of x in the following figure:
Inscribed Circle Problems
Solution:
 
Here x = Reflex∠AOB
From inscribed angle theorem,
Reflex∠AOB = 2∠ACB
Reflex∠AOB = 2 * 135°
Reflex∠AOB = 270°

The value of x is 270$^o$
 

Question 2: Find the value of x in the following figure:
Problem based on Inscribed Circle
Solution:
 
From inscribed angle theorem,
∠AOB = 2∠ACB
50° = 2 * ∠ACB
∠ACB = 25°
In ΔAOC:
OA = OC (radii)
Since, Angles opposite to equal sides are equal.
∴ ∠OAC = ∠OCA
∠x = 25° (as ∠OCA = ∠ACB)