Perpendicular from the Center to a Chord Bisects the Chord
Prove that di...
Question
Prove that diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.
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Solution
Given: AB and CD are two parallel chords of a circle with centre O. POQ is a diameter which is perpendicular to AB. To prove: PF ⊥ CD and CF = FD Proof: AB || CD and POQ is a diameter. ∠ PEB = 90° (Given) ∠ PFD = ∠ PEB (∵ AB || CD, Corresponding angles) Thus, PF ⊥ CD ∴ OF ⊥ CD We know that the perpendicular from the centre to a chord bisects the chord. i.e., CF = FD Hence, POQ is perpendicular to CD and bisects it.