To prove that AE/ED = BF/FC in trapezium ABCD with EF || AB, we can use similar triangles.

**Step 1: Identify Similar Triangles**
Let's draw lines parallel to AB passing through points E and F, intersecting DC at points G and H respectively.

We can observe that triangles ADE and GEF are similar, and triangles BCF and HFEG are similar. This is because:
- AE || GH (as EF || AB)
- AD || EG (as EG is parallel to AB)
- DE || FG (as FG is parallel to AB)
Similarly, we can conclude that:
- BF || GH
- HC || GH
- FC || HF

**Step 2: Establish Proportions**
Since triangles ADE and GEF are similar, we can write the proportion:
AE/ED = GE/EF ----(1)

Similarly, since triangles BCF and HFEG are similar, we can write the proportion:
BF/FC = HF/FE ----(2)

**Step 3: Equating EF and GE**
Since EF is parallel to AB, we can conclude that triangles GEF and GAB are similar. Therefore, we can write the proportion:
GE/EF = GA/AB

Since AB is parallel to DC, we can also conclude that triangles GAB and GDC are similar. Therefore, we can write the proportion:
GA/AB = GD/DC

Combining the above two proportions, we get:
GE/EF = GD/DC ----(3)

**Step 4: Equating HF and GD**
Since HF is parallel to AB, we can conclude that triangles HFEG and HAB are similar. Therefore, we can write the proportion:
HF/FE = HA/AB

Since AB is parallel to DC, we can also conclude that triangles HAB and HDC are similar. Therefore, we can write the proportion:
HA/AB = HD/DC

Combining the above two proportions, we get:
HF/FE = HD/DC ----(4)

**Step 5: Combining Proportions**
From equations (1), (3), and (4), we have:
AE/ED = GE/EF = GD/DC = HF/FE = BF/FC

Therefore, we can conclude that AE/ED = BF/FC in trapezium ABCD with EF || AB.

Since, AB II CD and AB II EF
therefore ,
AB II CD. II EF
join Ac which cut EF at point O

in triangle ACD ,
OE II DC
then by basic proportionality theorem ,
AE/ED = AO/OC. ..( i )

Similarly in triangle ABC
BF/FC = AO/OC. ..( ii )


on comparing equation ( i ) and ( ii ) we get ,
AE/ED = BF/ FC