Solution 1: Let the intersection of AB and CD be O, because ∠OAC=∠ODB, ∠AOC=∠DOB, so Δ δAOC?δDOB, so AO/CO=DO/BO. Because ∠AOD=∠COB, δ AOD δ COB, ∠ ADO =∠CBO, so ∠ ADE =∠ACB. Because AB=AC, ∠ABC=∠ACB, ∠ADO=∠ADE, so AD shares ∠EDC equally.

Solution 2: Because ∠BAC=∠BDC, A, C, B and D are * * * cycles, ∠ADC=∠ABC, ∠EDA=∠BCA. Because AB=AC, so ∠ABC=∠ACB, so ∠ADC=∠ADE, so AD shares ∠EDC equally.