Posted by qwerty123 on Monday, March 21, 2011 at 6:30pm.
extend AM its own length of 3.5 to point D.
Join BD and CD
You now have a quadrilateral where the diagonals AD and BC bisect each other.
So ABDC must be a parallelogram .
In triangle ADC you could now use the cosine law to find angle DAC
Once you have that you can use the cosine law again to find MC in triangle AMC
double MC to find BC
Triangle ABC, A at the top, B & C at the bottom, B left & C right.
The length of a triangle median is defined by
(Ma)^2 = (b^2 + c^2)/2 - a^2/4. (Ma is the median from vertex A to side "a".)
(Mb)^2 = (a^2 + c^2)/2 - b^2/4
(Mc)^2 = (a^2 + b^2)/2 - c^2/4
Letting BC = a, AB = c = 4, AC = b = 7 and AH = 3.5
Then, (3.5^2) = (7^2 + 4^2)/2 - a^2/4
Multiplying through by 4 yields 4(3.5)^2 = 2(49 + 16) - a^2 or
49 = 130 - a^2 or a^2 = 130 - 49 = 81 making a = 9 = BC.
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