In the figure , ABC is an isosceles triangle with AB=AC . The side CB is produced to D such that CB=BD . THe side AC is produced to E such that AC=CE . If EB is produced meets AD at X , prove that DX=XB .
let angles ABC = ACB = a
therefore angles ABD = BCE = $\displaystyle 180^o$ - a
and since DB = BC and AB = CE
you have congruent triangles ADB and EBC (SAS)
Hence, angles ADB = EBC
also, angles EBC = XBD (vertically opposite angles)
which means angles ADB = XBD = EBC
Hence, triangle XDB is isosceles, and DX = XB.