Q9 of 31 Page 65

The two circles with centers X and Y intersect each other at the points A and B. A is joined with the mid-point ‘S’ of XY and the perpendicular on SA through the point A is drawn which intersects the two circles at the points P and Q. Let us prove that PA = AQ.


Given, Two Circles with Centers X and Y intersect each other at point A and B. S is a mid-point of XY, AS is perpendicular to PQ.


To Prove: AP = AQ


Construction: XM perpendicular to chord PA. And YN perpendicular to chord AQ of respective circles.


Since SA is also perpendicular to PQ (given)


So, all these perpendiculars to the same line are parallel to each other.


SX = SY


AM = AN



So,


AP = AQ


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