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
