In the triangle ABC, ab = AC, EF is the median line of the triangle ABC, intersecting AB, AC at e, f respectively, extending AB to D, making BD = AB, connecting CD

In the triangle ABC, ab = AC, EF is the median line of the triangle ABC, intersecting AB, AC at e, f respectively, extending AB to D, making BD = AB, connecting CD

∵ AB = AC, EF is the median line of triangle ABC
∴AE=AB/2=AC/2=AF
∵AE=AF,∠A=∠A,AC=AB
∴△EAC≌△FAB
∴CE=BF
In triangular ADC, AF = FC, ab = BD
That is, B and F are the midpoint of AD and AC respectively
The BF is the median of the triangle ADC
∴BF=CD/2
∴CE=CD/2

In triangle ABC, ab = AC, EF is the median line of triangle ABC, extend AB to D, make BD = AB, connect CD, prove: CE = 1 / 2 CD

Connecting BF
According to the condition of the title, if the quadrilateral ebcf is isosceles trapezoid, then EC = BF
∵AB=BD,AF=FC,
Ψ BF is the median line of △ ADC
BF=½CD
∴EC=BF=½CD