+0

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,

0
946
2
+1760

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?

Mellie  Nov 20, 2015

#1
+78755
+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

CPhill  Nov 21, 2015
Sort:

#1
+78755
+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

CPhill  Nov 21, 2015
#2
+91053
+5

Another very nice answer Chris :)

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

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

Melody  Nov 22, 2015

3 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details