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April 25, 2014

April 25, 2014

Posted by **qwerty123** on Monday, March 21, 2011 at 6:30pm.

- Geometry -
**Reiny**, Monday, March 21, 2011 at 8:06pmextend 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

- Geometry -
**tchrwill**, Tuesday, March 22, 2011 at 9:36amTriangle 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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