Q16 of 62 Page 10

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

Let AB be diameter in a circle with center O and MN is the chord at point A.


And CD be any chord parallel to MN intersecting AB at E


To Prove: AB bisects CD.


Proof:


OA ⏊ MN [Tangent at a point on the circle is perpendicular to the radius through point of contact]


∠MAO = 90°


Also, ∠CEO = 90°


[ if E point lies on the OA then by Corresponding angles and if E lies on OB then by sum of co-interior angles]


So, we have OE ⏊ CD


CE = ED [ Perpendicular drawn from the center of a circle to chord bisects the chord]


Implies that AB bisects CD


Hence Proved!


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