Find the length of the median AE of triangle ABC with sides AB=8, BC=6, and angle bisector BD=6

Answer 1

By property of angle bisector AD/DC = 8/6 = 4/3;

By cosine law cos(B/2) = (8² + 6² - AD²)/(2·8·6) = (6² + 6² - DC²)/(2·6·6); (100 - AD²)/96 = (72 - DC²)/72;

3(100 - AD²) = 4(72 - DC²); 300 - 3AD² = 288 - 4DC²; 3AD² - 4DC² = 12; 3AD = 4DC; DC = 3AD/4;

3AD² - 4(9AD²/16) = 12; 3AD² - 9AD²/4 = 12; 12AD² - 9AD² = 48; AD² = 16; AD = 4; DC = 3. SO, AC = 7.

Now, if we continue median AE and take EF + AE we get parallelogram ABFC. Using property of sides and diagonals of parallelogram we get: (2AE)² + 6² = 2·8² + 2·7²; AE² = ((128 + 98 - 36)/4 = 190/4;

AE = √190/2