I will assume that D is the midpoint of AB
draw the median from A to BC to meet BC at E
let the intersection of these two medians be F
Since you have an isosceles triangle that median will meet BC at right angles.
Also the medians intersect each other in the ratio of 2:1, the longer side towards the vertex.
Then FC = 6
by Pythagoras
FE^2 + 16 = 36
FE = √20
then AE = 3√20
finally AC^2 = AE^2 + EC^2
= 180 + 16
= 196
AC = √196 = 14
But AB = AC
so AB = 14
what is the length of side AB in triangle ABC with AB=AC, BC=8, and median CD=9
1 answer