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07272007  #1 
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plzzzzzzzzzzzzz !!!!Help!!!me solve dis geometry question.....?
im givin link for wich has the figure of the triangle question...plzz open it...http://www.freewebs.com/dandiwal/triangl...Question is is mid point of side BC of triangle ABC and ray AX is drawn as shown in figure.BM,CN and DG are perpendiculars to AX in M, N, G respectively. Prove that MG=GN...just give da reason...

07272007  #2 
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your link did not work. do something about it.Okay: i saw the picture.Let AX intersect BC at T.1) Triangles TDG and TBM are similar triangles TDG and TCN are similar (AA postulate in each case)2) DG // BM..................(two lines perpendicular to same line //)3) TG/ GM = TD/ BD....if a line is // to one side of triangle, it will divide other sides proportionally)4) TN/TG = TC/TD (sides of similar triangles are proportional)5) (TN + TG)/TG = (TC +TD)/ TD ...(property of proportion)6) GN/ TG = CD/TD.... (substitution, since TN + TG = NG)7) TG/GN = TD/ CD.....(property of proportion)8) TG/GN = TD/ BD... (substitution, since BD=CD)9) TG/GN = TG/GM... (substitution steps 3 and 8)10) GN = GM ......algebra

07272007  #3 
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The link that you provided is incorrect.

07272007  #4 
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sorry, i wanted to answer it but your link doesn't work. i tried to draw it myself... but then i got confused.edit it...?!

07272007  #5 
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You should use Thales theorem (not the one about right triangle in circle, but the one about parallel line to a triangle side).BM, CN and DG are all parallels each other, because they're all perpendicular to a same straight line.As D is BC mid point, we have BD = DCor we can also say (BC/BD) = 2 and (BC/DC) = 2And thus, if BD = DC with BM // DG // CN and with M, G and N on the same lineWe have a ratio conservation ; that means (MN/MG) = 2 and also (MN/GN) = 2thus we find that MG = GN;;;
en.wikipedia.org 
07272007  #6 
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link is not working

07272007  #7 
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Doesn't seem as though my first response came through.OK, this is going to be tough to explain via words, but here we go. First of all the diagram you have is tough to look at, redraw it such that D is actually the midpoint of BC and ray AX is 90 deg to BM, GD and CN. Now draw a new line that intersects line BC at D and is also perpendicular to BM, GD and CN. Give this line end points Z1 and Z2. Now you've just created two new right triangles (Z1BD and Z2DC) which have exactly equal hypotenuses (BD = DC, because D is the midpoint of BC). Right triangles with exactly equal hypotenuses means that all the other sides must be equal as well. Therefore your new lines Z1D = DZ2 and Z1D = MG and DZ2 = GN...and you can figure it out from there.Take care and keep having fun.

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