I don't remember the names of the theorems at all, but we know that AD ≈ DE ≈ EB (Segment Addition). Which means <ACD ≈ <DCE ≈ <ECB (not completely sure how to prove this). Therefore, <1 ≈ <2. (Don't forget to state that <1 ≈ <ACD and <2 ≈ <ECB). With this and Angle Addition, we know that <ACE ≈ <DCB, so with SAS Congruence Theorem, we can prove ΔACE ≈ ΔBCD.
