Given: ∆abc, ab = cb bd − median to ac e∈ ab ,f∈ bc ae = cf prove: △ade ≅ △cdf δbde ≅ δbdf

1) By SAS theorem, ΔADE≅ΔCDF

2) By SSS theorem, ΔBDE≅ΔBDF

Step-by-step explanation:

Consider isosceles triangle ABC (see diagram).

1. In triangles ADE and CDF:

AD≅DC (since BD is median, then it divides side AC in two congruent parts);AE≅CF (given);∠A≅∠C (triangle ABC is isosceles, then angles adjacent to the base are congruent).

By SAS theorem, ΔADE≅ΔCDF.

2. In triangles BDE and BDF:

side BD is common;DE≅DF (ΔADE≅ΔCDF, then congruent triangles have congruent corresponding sides);BE≅FB (triangle ABC is isosceles, AB≅BC, AE≅CF, then BE=AB-AE, FB=BC-CF).

Be SSS theorem, ΔBDE≅ΔBDF.