Let M be the midpoint of side AB of triangle ABC. Angle bisector of AD of angle CAB and the perpendicular bisector of side AB meet at X. If AB=40 and MX=9, then how far is X from line AC?
Here is what I believe you describe :
AD bisects angle CAB, CM is the perpendicular bisector of AB
XM = 9, AB = 40
And triangle AMX ≈ triangle AEX .....so..... AX/MX = AX/EX .......so......MX = EX ....and EX is the distance from X to AC = 9