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Math problem of grade three-similar triangles. Master, come quickly.
Solution: connect AE;

∵EF⊥ Divide advertisements equally;

∴ae=de;

∴∠dae=∠ade;

∫∠BAD =∠CAD; ;

∴∠bad+∠eac=∠cad+∠eac=∠dae=∠ade;

∵∠b+∠bad =∠ade;

∴∠bad+∠eac=∠b+∠bad;

∴∠eac=∠b;

∠∠EAC =∠B;

∠CEA =∠AEB;

∴△ace∽△bae;

∴ae:ce=be:ae;

∴AE? = CE * BE

ed = AE

∴ED? = CE * BE