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triangles
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 .

You have to prove that the triangles are similar. But this is obviously because AB ia parallel to DZ.