Question

Given: ∆ABC, AB = CB BD − median to AC E∈ AB ,F∈ BC AE = CF Prove: △ADE ≅ △CDF ΔBDE ≅ ΔBDF

Answer 1

Answer:

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.