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?

beastrohit Nov 14, 2018

#1**+1 **

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

CPhill Nov 14, 2018