The median BD is drawn in an isosceles triangle ABC with base AC.

The median BD is drawn in an isosceles triangle ABC with base AC. points E and F are marked on sides AB and CB, respectively, so that AE = CF. prove that BDE = BDF, ADE = CDF.

Consider all the resulting triangles.

In a triangle ABC AB = BC, as the sides of an isosceles triangle, the median BD is the height and the bisector, EB = AB – AE = BF = BC – FC. Angles <ABD = <DBC; <BAD = BCA.

Then the triangles EBD; BFD equal on equal sides EB = BF; BD is common, and the angle between them is <EBD = <DBF.

Triangles AED; DFCs are also equal on both sides and the angle between them:

AE = FC (by condition); AD = DC since DB is the median; <BAC = <BCA, as base angles in an isosceles triangle ABC.

Proven.