Prove that the line joining mid points of 2 equal chords of a circle subtand equal angles with the chord
2 circles have AB as their common chord AC is diameter AD is diameter of second circle
show B,C,D are collinear

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See diagram.

1st problem

the two chords must be adjacent. One point is common between two chords.  Then only it is true.  We take angles subtended at the center O of the circle.  The angles at the circumference also will be proved similarly.

angle AOB = angle BOC   as AB and BC are of same length. 

angle AOD = 1/2 angle AOB : isosceles triangle and D is midpoint.

angle COE = 1/2 angle BOC : isosceles triangle  and E is midpoint.

Now the angle DOE = angle DOB + angle BOE = angle AOB  =  angle BOC.

2nd problem

    AB is the common chord.  AC is the diameter of the 1st circle.  Then the angle ABC is the angle subtended by the diameter AC at the point B on the circumference. Thus it is 90 deg.

    Similarly, in the 2nd circle, the diameter AD subtends angle ABD at the point on the circumference.  So it is 90 deg.

   The sum of the angles ABC and ABD at point B is 180 deg.  So CBD is a straight line, the angle at B is equal to straightline angle.

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reply sir
i didnot understand anything in ques 2
can you explain