Answer:
Given : In ∆ABC,
AB = CB, BD is median to AC, E∈ AB, F∈BC, AE = CF
To prove : Δ ADE ≅ Δ CDF and Δ BDE ≅ Δ BDF
Proof :
Join C and F to the point D. ( construction )
(i) In triangles ADE and CDF,
AE = CF ( given )
AD = CD ( BD is median to AC )
m∠EAD = m∠FCD,
By SAS postulate of congruence,
ΔADE ≅ ΔCDF
By CPCTC,
DE = DF
(ii) In triangles BDE and BDF,
BD = BD ( common segments ),
DE = DF,
EB = FB ( ∵ AB = BC ⇒ AE + EB = CF + FB ⇒ AE + EB = AE + FB ⇒ EB = FB )
By SSS congruence postulate,
ΔBDE ≅ ΔBDF
Hence, proved........
