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AlexandervonHumboldt2
 one year ago
Geometry Question.
AlexandervonHumboldt2
 one year ago
Geometry Question.

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AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.1Side 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

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.1@ganeshie8

AlexandervonHumboldt2
 one year ago
Best ResponseYou've already chosen the best response.1Here 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+164x=(2*sqrt19)^2 using this equation find that x=10, thus AC=10. i'm i correct @ganeshie8 ?
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