Alternate Segment Theorem

Given a circle and two points P and Q on its boundary and a departure to the circle at Q, then angle between the departure and the line PQ is the same as the angle subtended by this chord in segment of the circle on the opposite side of PQ.  Alternate segment theorem states that approach involving a departure and its chord is equal to angle in the alternate segment.Interactive animatronics representatives proof of an alternate segment theorem.

Theorem for alternative segment

         The length of the line segment AB, which joins A (x₁, y₁) and B (x₂, y₂) is given by
d= `|AB|=sqrt((x2-x1)^(2)+(y2-y1)^(2))`  
Proof:
Let A (x₁, y₁) and B (x₂, y₂) be two points in the plane.
Let d = distance between the points A and B.
Draw AL and BM perpendicular to x-axis (parallel to y-axis).
Draw AC perpendicular to BM to cut BM at C.

OL = x₁, OM = x₂ [AC = LM = OM - OL = x₂ - x₁]
MB = y₂, MC = LA = y₁ [CB = MB - MC = y₂ - y₁]
`AB^(2)=AC^(2)+CB^(2)`
`d^(2)=(x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2)`
d=`sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1)^(2)))`

Alternate Segment Circle Theorem

           The chord CE divides the circle into 2 segments. Angle CEA and CDE are angles in alternate segments because they are in opposite segments. Circles are easy blocked curves which separate the plane into two regions, an interior and an exterior.The alternating segment theorem states that a position connecting a departure and a chord through the point of contact is equal to the angle in the alternate segment.
         In conditions of the beyond diagram, the alternating segment theorem tells us to facilitate angle CEA and angle CDE are equivalent.



A digression makes an angle of 90 degrees with the radius of a circle,so we know that ∠OAC + x = 90.
The angle in a semi-circle is 90, so ∠BCA = 90.
The angles in a triangle add up to 180, so ∠BCA + ∠OAC + y = 180.
Therefore 90 + ∠OAC + y = 180 and so ∠OAC + y = 90.
But OAC + x = 90, so ∠OAC + x = ∠OAC + y.
Hence x = y.