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Circles in Euclidean

The

There’s a bunch of them. The main thing of theoretical note is that these results exist only in . Again, proofs in text, focus on computation.

Inscribed The measure of an 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 .

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 and a 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: