E and D are mid points of BC and AC respectively, \$BD \bot AC\$, \$AE \bot BC\$ and AB = BC = CA, then prove \$\triangle BDC \cong \triangle AEC\$.

\$ED \bot AC\$, B is the mid-point of ED and AC. Prove \$\triangle EBC \cong \triangle DBA\$.

\$\angleB = \angleE = 90^\circ\$, AC = DF, E and B are the mid-point of DF and AC respectively, and AE = FB. Prove \$\triangle EBA \cong \BEF\$.

\$\angleA = \angleC = 90^\circ\$, B is the mid-point of AC and DE. Then \$\triangle DBA \cong \triangle EBC\$.

\$\angleE = \angleC = \angleF = 90^\circ\$, B is the mid-point of AC and DG respectively. Prove \$\triangle GCB \cong \triangle BEF\$.