Given: ∆abc, ab = cb

bd− median to ac

e∈ ab, f∈bc

ae = cf

prove: △ade ≅ △cdf

δbde ≅ δbdf

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........

b

step-by-step explanation: