chords equidistant from the centres of a circle are eqal in length
Answers
Theorem: Congruent chords of circle are equidistant from the center of the circle.
Given: 'O' is the center of the circle, where Chord AB ≅ Chord MN.
To prove that: CO ≅ PO
Construction: Draw radii OB and radii ON.
Proof: OC ⊥ AB and OP ⊥ MN (Given)
Therefore,
Seg AB = Seg PN (Given)
CB = 1/2 AB; PN = 1/2 MN
∴ Seg CB = Seg PN ------ (i)
Now,
In ΔOCB and ΔOPN,
Seg CB ≅ Seg PN ----- From i
∠OCB ≅ ∠OPN ----- Each 90°
Seg OB ≅ Seg ON ------- Radii of circle
∴ ΔOCB ≅ ΔOPN (Hypo. side test)
∴ Seg OP = Seg CO ------- C.S.C.T
Therefore, Chords are equidistant from the center of the circle.
"Refer to the Given attachment".
Step-by-step explanation:
Theorem: Congruent chords of circle are equidistant from the center of the circle.
Given: 'O' is the center of the circle, where Chord AB ≅ Chord MN.
To prove that: CO ≅ PO
Construction: Draw radii OB and radii ON.
Proof: OC ⊥ AB and OP ⊥ MN (Given)
Therefore,
Seg AB = Seg PN (Given)
CB = 1/2 AB; PN = 1/2 MN
∴ Seg CB = Seg PN ------ (i)
Now,
In ΔOCB and ΔOPN,
Seg CB ≅ Seg PN ----- From i
∠OCB ≅ ∠OPN ----- Each 90°
Seg OB ≅ Seg ON ------- Radii of circle
∴ ΔOCB ≅ ΔOPN (Hypo. side test)
∴ Seg OP = Seg CO ------- C.S.C.T
Therefore, Chords are equidistant from the center of the circle.
"Refer to the Given attachment".
