two circles with centres P and Q intersect at C and D. PM=QM, AB perpendicular CM. Prove that AC=BC​

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:

hello from mrbilldoesmath!

answer:

the domain is the set { -2, 0, 2, 3}

discussion:

the domain is the set of x values in the ordered pairs (x,y). taking the x, or first component, of each ordered pair gives   -2, 0, 2, 3.

you,

mrb

answer: idk wbu

step-by-step explanation: umm..