Theorem 7

Transcript

Theorem 7 Chords of a circle that are equidistant from the centre have equal length Given: A circle with center at O; AB and CD are two Chords of the circle where OX is distance of chord AB from center i.e. OX ⊥ AB; OY is distance of chord AB from center i.e. OY ⊥ CD; & both distances are equal OX = OY To Prove: AB = CD Proof: We make ∆ AOX & ∆ COY congruent, and by CPCT we can prove AX = CY, which are half of AB & CD In ∆AOX and ∆COY ∠OXA = ∠OYC OA = OC OX = OY ∴ ∆AOX ≅ ∆COY Thus, by Corresponding parts of Congruent Triangles (CPCT) AX = CY We know that Perpendicular from the center to the chord, bisects the chord For Chord AB Since OX ⊥ AB, ∴ X bisects AB. So, we can write AB = 2AX For Chord CD OY ⊥ CD ∴ Y bisects CD. So, we can write CD = 2CY From (1) AX = CY Multiplying by 2 both sides 2AX = 2CY Putting 2AX = AB & 2CY = CD from (2) &(3) AB = CD Hence proved