By a similar argument used to prove that △AEB ≅ △CED, we can show that △ ≅ △CEB by. So, ∠CAD ≅ ∠ by CPCTC. Therefore, AD || BC by the converse of the theorem. Since both pair of opposite sides are par:
AED
By a similar argument used to prove that △AEB ≅ △CED, we can show that △ ≅ △CEB by. So, ∠CAD ≅ ∠ by CPCTC. Therefore, AD || BC by the converse of the theorem. Since both pair of opposite sides are par:
SAS
By a similar argument used to prove that △AEB ≅ △CED, we can show that △ ≅ △CEB by. So, ∠CAD ≅ ∠ by CPCTC. Therefore, AD || BC by the converse of the theorem. Since both pair of opposite sides are par:
ACB
By a similar argument used to prove that △AEB ≅ △CED, we can show that △ ≅ △CEB by. So, ∠CAD ≅ ∠ by CPCTC. Therefore, AD || BC by the converse of the theorem. Since both pair of opposite sides are par: