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Prove that a diameter AB of a circle bisects all those chords which are parallel to the tangent at the point A.

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Solution
According to question
     

Let us take a chord EF || XY
Here    \angle XAO= 90^{\circ} 
(\because tangent at any point of the circle is perpendicular to the radius through the point of contact)

\angle EGO= \angle XAO   (Corresponding angles)

\therefore \, \angle EGO= 90^{\circ}
Thus AB bisects EF
Hence AB bisects all the chords which are parallel to the tangent at the point A.
Hence Proved

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