Question

10. Prove that the 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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Answer 1

Answer: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.

Step-by-step explanation: