Q. 104.0( 24 Votes )
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 .
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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