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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
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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   

 

 

 

cool cool cool

CPhill  Nov 14, 2018
edited by CPhill  Nov 14, 2018

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