In a Δ ABC, BD is the median to the side AC, BD is produced to E such that BD = DE.
Prove that : AE is parallel to BC.
In the given △ABC ,
BD is a median
and
BD = DE
In △BCDand△EAD
CD = DA [BD is median]
BD = DE [given]
∠BDC=∠EDA [vertically opposite angle]
By SAS,
△BCD≅to△EAD
Then by cpctc
\(\angle BCD = \angle EAD \)
→ when angles are equal then
AE is ∥ BC .