Altitude of a circle!
Statement 1: The point of intersection of common chords of three circles described on the three side of a triangle as diameter is orthocenter of the triangle.
Statement 2: The common chords of three circles taken two at a time are altitudes of the triangle.
NOTE:- This is not an original question.
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1 solution
This problem is basically a coordinate geometry problem and kind of the one faced during preparation of JEE advanced exam in India. However one can also approach this via geometry. I think via coordinate this is done quickly. Let's assume vertices of the triangle be Pi=(xi,yi), for i=1,2,3. Now equation of a circle passing through P1 and P2 making them the diametric end points is (x-x1)(x-x2)+(y-y1)(y-y2)=0. Similarly eqn of circle through P1 and P3 is (x-x1)(x-x3)+(y-y1)(y-y3)=0. Now eqn of common chord of 2 circles is simply obtained by subtracting their equations. Thus the common of the 2 circles written earlier is (x-x1)(x2-x3)+(y-y1)(y2-y3)=0. On rearranging eqn of this line a bit(point-slope form) it becomes (y-y1)/(x-x1)=-(x2 -x3)/(y2 -y3). Clearly it represents the eqn of line passing through P1 and has a slope perpendicular to that of line P2P3, which is essentially the altitude of triangle. Thus second statement is true and the first statement follows.