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In two triangle abc and triangle def. if ab,bc and median ax are respectively proportional to df ,ef and median dy . then prove that triangle abc- triangle



Ax and Dy are medians, therefore BX is proportional to FY both being respectively half of proportional sides BC and FE . Therefore triangles ABX and DFY are similar (all sides are proportional). Hence angle AXB in triangle ABX = angle DYF in triangle DFY   -----(1).   Now in triangle AXC, angle AXC = 180 - angle AXB, and in triangle DEY, angle DYE = 180 - angle DYF ---------------(2). From (1) and (2), Angle AXC in triangle AXC = Angle  DYE in triangle DYE. ------------(3)                                                                    Now consider triangles AXC and DYE. sides Xc and YE are proportional , each being half of proportional sides BC and EF ----------(4)                     Sides AX and DY are proportional (being given proportional medians)---(4)                  Included angle AXC = Included angle DYE (from equation (3) above  ----(5). Hence triangles AXC and DYE are similar. Therefore sides AC and DE in triangles AXC and DYE respectively are proportional -----(6).                               Now in triangles ABC and DFE, sides AB, BC are given to be proportional to sides DF and FE,and from (6) the third side AC is proportional to DE.  Hence all sides are proportional. Therefore triangles ABC and DFE are similar.  
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