Let M be the midpoint of side AB of the triangle ABC. Angle bisector 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?
MA = 20 and MX = 9
And the perpendicular bisector of AB = CM
So, triangle MAX will be a right triangle with angle AMX = 90°
Draw XN perpendicular to AC
And AD bisects angle BAC...so....
And angle MAX = angle NAX
And angle AMX = angle ANX = 90°
And AX is a common side to triangles AMX and ANX
So....by AAS, triangle AMX is congruent to triangle ANX
So.....MX = NX = 9 = the distance from X to AC