2 1. As shown in the figure, DE is the center line of △ABC, ∠ B = 90o, AF‖BC. Is there a point m on the ray AF that makes △MEC similar to △ADE? If it exists, please determine the point m first, and then prove that the two triangles are similar; If it does not exist, please explain why.

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Solution: existence, the intersection e is EM⊥AC, AF is at point m, then m is. Proof: link MC. Em⊥acae = EC∴∠Mae =∠MCE \af BC ∴∠ Mae = ∠ ACB ∴∠.