Answer:
Two circles with centres P and Q intersect in C and D. M is the midpoint of PQ. Through C, a straight line ACB is drawn perpendicular to CM to meet the circles at A and B. How can you prove that AC=BC?
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Let L and N be the foot of the perpendiculars from P and Q to ACB.
PL,,MC and QN are parallel lines and make equal intercepts on line PQ.
Like wise, the intercepts on lineACB will also be equal. viz.
NC=CL
ButLand N are foot of perpendiculars from centers of circles to the corresponding chords . So L and N bisect chords AC and CB. Hence AC=BC
Step-by-step explanation:
