Circles in Euclidean Geometry

Circles in Euclidean Geometry

Circles in Euclidean Geometry The circle theorems There’s a bunch of them. The main thing of theoretical note is that these results exist only in Euclidean geometry. Again, proofs in text, focus on computation. Inscribed angle theorem The measure of an inscribed angle of a circle equals one half that of its intercepted arc. 1 m∠ABC = mAC 2 Example: Given mAC = 80, find m∠ABC. Solution: Corollary 1 An angle inscribed in a semicircle is a right angle. AD is a diameter. m∠ABD =90 m∠ACD =90 Corollary 2 An angle whose vertex lies inside a circle and is formed by intersecting chords of the circle x y 1 x y (intercepting arcs of measures and ) has measure 2( + ). 1 m∠BAC = (mBC + mDE) 2 1 m∠DAE = (mBC + mDE) 2 Example: Given mBC =68andm∠BAC = 60, find y = mED. Solution: Corollary 3 An angle whose vertex is exterior to a circle and is formed by intersecting secants of the x y 1 x − y circle (intercepting arcs of measures and ) has measure 2( ). 1 m∠DAE = (mBC − mDE) 2 Example: Given mED =32andm∠EAD = 18, find x = mBC. Solution: Corollary 4 An angle formed by a chord and a tangent of a circle, with its vertex at the point of x 1x tangency and intercepting an arc of measure on that circle has measure 2 . 1 m∠BAC = mAB 2 Example: Given m∠BAC = 50, find x = mBA. Solution: Two chord theorem When two chords of a circle intersect, the products of the lengths of the segments formed on one chord equals that on the other chord. AP · BP = CP · DP Example: Given CP = 50, DP = 30, BP = 45, find x = AP . Solution: Secant tangent theorem ←→ ←→ If a secant PA and a tangent PC meet a circle at the respective points A, B,andC (point of tangency), then (CP)2 = AP · BP. (CP)2 = AP · BP Example: Given AP = 40, BA = 100, find x = CP. Solution: Two secant theorem ←→ ←→ If two secants PA and PC of a circle meet the circle at A, B, C,andD,thenPA· PB = PC · PD PA· PB = PC · PD Example: Given AP = 40, BA = 120, CP = 50, find x = CD. Solution:.

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