PRINCIPIA · THEOREM

The central angle equals twice the inscribed angle on the same arc

Dependencies: Base angles of an isoceles triangle are equal (base angles of an isosceles triangle are equal), Exterior angle equals the sum of two non-adjacent interior angles (an exterior angle equals the sum of the two non-adjacent interior angles).

Statement

Let AA, BB, PP be three points on ⊙O\odot O, with PP not on the segment of line ABAB between AA and BB (i.e. PP lies on the "far-side" arc subtended by chord ABAB and OO). ∠APB\angle APB is called the inscribed angle subtending arc ABAB, and ∠AOB\angle AOB the central angle subtending the same arc. Then

∠AOB  =  2 ∠APB.\angle AOB \;=\; 2\,\angle APB.

In other words, the inscribed angle is exactly half the central angle on the same arc.

On circle O: the inscribed angle \angle APB = \alpha and the central angle \angle AOB = 2\alpha on the same arc.

10 foundational theorems free · one-time unlock for the rest

Unlock the complete proofs forever

Matching GeoSnap, 10 foundational theorems are completely free. Unlock every other proof, animation, and consequence with one $10 purchase.

Sign in to unlock
Help me make this theorem better