In triangle ABC, ab = AC, EF is the median line of triangle ABC, extend AB to point D, make BD = AB, connect CD, what do you think is the relationship between CE and CD

In triangle ABC, ab = AC, EF is the median line of triangle ABC, extend AB to point D, make BD = AB, connect CD, what do you think is the relationship between CE and CD

1:2
Because AE / AC = AC / ad = 1 / 2, triangle AEC is similar to triangle ACD,
So CE / CD = 1 / 2

In the triangle ABC, ab = AC, EF is the median line of △ ABC, intersecting AB, AC at e, f respectively, extending AB to D, making BD = AB connect CD, proving: CE = 1 / 2ce
Don't use similarity

∵ EF is the median line of △ ABC and intersects AB and AC with E and f respectively
∴EF∥BC,EF=1/2BC
∴∠EFC=∠CBD
∵AB=AC=BD,FC=1/2AC
∴FC=1/2BD
According to the triangle similarity theorem
△EFC≌△DBC
∴CE=1/2DC

As shown in the figure, De is the median line of triangle ABC, point F is on De, and

The EF length is 3

As shown in the figure: De is the median line of △ ABC, AF is the median line on the side of BC, and de and AF intersect at point o
Verification: De and AF are equally divided

Connecting DF and ef
Because De is the median line and F is the midpoint
So DF parallel AE EF parallel ad
So ADFE is a parallelogram
So de and AF are bisected (the diagonals of parallelogram are bisected)