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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Side BC of triangle ABC is 4, side AB is 2*sqrt(19). Given that the center of the circle, passong though midpoints of sides of triangle, lies on the bisector of angle C. Find AC
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looks tricky...
Here is what I have done so far: Let A_1, B_1, C_1 are the midpoints of BC, AC, AB. O is the center of circle. /_ACB=a. . As angle A_1C_1B_1 is congruent to angle ACB and is congruent to a, thus triangle A_1B_1C_1 is equal to triangle B_1A_1C, Consequently raduises of given circle and circle circumscribed around triangle A_1B_1C are equal. Let line OC intersects the second circle at point M. MA_1=MB_1. OA_1=OB_1. If O and M are not same points, then OC _|_A_1B_1. CA_1=CB_1, AC=BC=4. in this case AC+BC=4+4=8 < 2*sqrt19=AB which is impossible. Thus M and O are same points. Thus anle A_1OB_1 + angle A_1CB_1 = 180°. thus a=60. AC=x, by theorem of cosines x^2+16-4x=(2*sqrt19)^2 using this equation find that x=10, thus AC=10. i'm i correct @ganeshie8 ?
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