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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