Answer:
see below
Step-by-step explanation:
2. ΔCDE is isosceles so ∠CDE ≅ ∠DCE
3. ∠CDE ≅ ∠EBA (alternate interior angles)
4. ∠DCE ≅ ∠EAB (alternate interior angles)
5. based on #3 and #4 =>∠EBA ≅ ∠EAB
6. because ∠EBA ≅ ∠EAB then ΔEBA is isosceles EA ≅ ∠EB
7. since ΔEBA and ΔCDE are isosceles, EA ≅ EB and ED ≅ EC
AC ≅ AE + EC
BD ≅ BE + ED
then ∠AC ≅ ∠BD
8. DC is the same for both ΔACD and ΔBDC
9. We have
∠CDE ≅ ∠DCE (#2)
AC ≅ BD (#7)
DC ≅ DC (#8)
10. Using Side-Angle-Side theorem => ΔACD ≅ ΔBDC
