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AlexandervonHumboldt2

  • one year ago

Geometry Question.

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  1. AlexandervonHumboldt2
    • one year ago
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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

  2. AlexandervonHumboldt2
    • one year ago
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    @ganeshie8

  3. ganeshie8
    • one year ago
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    .

  4. ganeshie8
    • one year ago
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    looks tricky...

  5. AlexandervonHumboldt2
    • one year ago
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    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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