In ΔABC, m∠CAB = 60° and

AD

is an angle bisector with D ∈

BC

and AD = 8 ft. Find the distances from D to the sides of the triangle.

The answer is 4 feet.

Draw a line segment from D perpendicular to AC and intersect AC at F.

triangle ADF is a right triangle, with angle AFD=90, Angle DAF=half of BAC=30

in a 30-60-90 triangle, the shortest side is half of the hypotenuse. the hypotenuse AD=8, so DF=4

Draw a line segment perpendicular from D to AB and intercept AB at E.

DE is also 4 feet, same as above.