Question

Given: BD bisects AC and ∠CBE≅∠ADE. Prove: ABCD is a parallelogram.

Answer 1

Using supplementary angle theorem

• <CDE=<ABE

As BD bisects AC

• AE=EC
• BE=ED

So ∆AEC$\cong$∆DEC(SAS)

Also similarly

• ∆BEC$\cong$∆AED

Hence it's a parallelogram

Answer 2

Answer:

Using supplementary angle theorem

<CDE=<ABE

As BD bisects AC

AE=EC

BE=ED

So ∆AEC∆DEC(SAS)

Also similarly

∆BEC∆AED

Hence it's a parallelogram