ABCD is parallelogram where X is a point that lies on BD such that DX=3XB . AX extended meets BC at Y and DC is extebded to Z . Prove that triangle AXB is similar to triangle ZXD .

Since ABCD is a prarallelogram , angle BAX = angle XZD .

angle ABX = angle XDZ

Then how can i make use of this information , DX=3XB to prove .

$\displaystyle DX\neq XB$ . But it's proportionate . So if the sides are proportionate , can i say that these triangles are congruent .