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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?

 Nov 20, 2015

Best Answer 

 #1
avatar+111433 
+20

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

 

 

 

 

 

 

 


 

 

 

 

cool cool cool

 Nov 21, 2015
 #1
avatar+111433 
+20
Best Answer

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

 

 

 

 

 

 

 


 

 

 

 

cool cool cool

CPhill Nov 21, 2015
 #2
avatar+110279 
+5

Another very nice answer Chris :)

You are becoming an expert at using GeoGebra    -:))

 

these geometry questions that Mellie is presenting are really fun. :)

 Nov 22, 2015

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