Question

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

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.

Hope you like my answer.

Answer 2

Answer:

hello friends,

Step-by-step explanation:

let AB and CD be the two parallel chords of the circle with center O. POQ is a diameter which is perpendicular too AB.

To prove: PF Perpendicular to CD and CF

perpendicular to FD.

proof:AB||CD and POQ is a diameter.

angle PEB=90°(given)

angle PFD=angle PED

thus,PF perpendicular to CF

so OF is perpendicular to CD

we know that the perpendicular form the

center to a chord bisect the chord.

i.e, CF perpendicular to FD

hence, POQ is perpendicular to CD and

bisect it.