In triangle ABC we have that AB=AC=14 and BC=20. What is the length of the shortest angle bisector in ABC? Express your answer in simplest radical form.
This triangle is isosceles....the shortest angle bisector is that of angle BAC =
sqrt [ AC^2 - (BC/2)^2 ] =
sqrt [ 14^2 - 10^2 ] =
sqrt 96 =
4 sqrt 6
