# Prove that OPP sides of a quadrilateral circumscribing a circle subtend 180 degree angle at the center of the circle

Let ABCD be a quadrilateral circumscribing a circle centered at O such that it touches the circle at the point P, Q, R, and S.

From Fig, in ∆ OAP and ∆ OAS,

AP = AS (Since they are tangents from the same point)

OP = OS (Radii of same circle)

OA = OA (Common side)

Therefore, by SAS criterion, ∆ OAP and ∆ OAS are congruent.

Hence, By C.P.C.T., ∠ POA = ∠ AOS or ∠ 1 = ∠ 2

Similarly, we can get, ∠ 2 = ∠ 3, ∠ 4 = ∠ 5, ∠ 6 = ∠ 7

Now, we know that total angle around a point is 360ᵒ

So, ∠ 1 + ∠ 2 + ∠ 3 + ∠ 4 + ∠ 5 + ∠ 6 + ∠ 7 + ∠ 8 = 360ᵒ

(∠ 1 + ∠ 8) + (∠ 2 + ∠ 3) + (∠ 4 + ∠ 5) + (∠ 6 + ∠ 7) = 360ᵒ

2∠ 1 + 2∠ 2 + 2∠ 5 + 2∠ 6 = 360ᵒ

2(∠ 1 + ∠ 2) + 2(∠ 5 + ∠ 6) = 360ᵒ

(∠ 1 + ∠ 2) + (∠ 5 + ∠ 6) = 180ᵒ

∠ AOB + ∠ COD = 180ᵒ

In similar way, it can be proved that ∠ BOC + ∠ DOA = 180ᵒ

Hence, opposite sides of a quadrilateral circumscribing a circle subtend 180 degree angle at the center of the circle.