Plane Geometry - Bruce E. Shapiro

SECTION 41. EUCLIDEAN CIRCLES 221Figure 41.2: Proof of theorem 41.3.Theorem 41.5 (30-60-90 Theorem) If the interior angles of triangle△ABC measure 30, 60, and 90, then the length of the side opposite the 30angle is one half the length of the hypotenuse.Proof. (Exercise.)Theorem 41.6 (Converse to 30-60-90 Theorem) If △ABC is a righttriangle such that the length of one leg is one-half the length of the hypotenusethen the interior angles of the triangle measure 30, 60, and 90.Proof. (Exercise.)Definition 41.7 Let Γ be a circle. An angle ∠P QR is said to be inscribedin Γ if P, Q, R ∈ Γ. The arc intercepted by ∠P QR is the set of all pointsin Γ in the interior of ∠P QR.Definition 41.8 Let Γ = C(O, r) be a circle. An angle ∠P OR (where Ois the center of Γ), is called a central angle.Definition 41.9 Let ∠P QR be inscribed in Γ = C(O, r) such that Q, R lieon opposite sides of ←→ OP or O, P lie on opposite sides of ←→ QR. Then ∠P ORis called the corresponding central angle.Theorem 41.10 (Central Angle Theorem) The measure of an inscribedangle for a circle is one half the measure of the corresponding central angle.Proof. Let ∠P QR be inscribed in Γ = C(O, r). We will consider threecases: (1) O lies on ∠P QR; (2) O lies in the interior of ∠P QR; (3) O liesin the exterior of ∠P QR.Revised: 18 Nov 2012 « CC BY-NC-ND 3.0.