CBSE Class 10 Answered

Note that,

In triangles DEA and FEB,

angle DEA=angle FEB(vertically opposite angles)

angle EDA=angle EFB( alternate int angles , with DA parallel to CB i.e. CF)

So,

 triangle DEA similar to triangle FEB (AA rule)

DA/EA=FB/EB...(i)

Next,

 in triangles FEB and FDC,

angle F is common

angle FEB = angle FDC(as EB and DC are parallel, so corr angles are equal)

so,

 triangle FEB similar to triangle FCD.

sO,

 FB/EB=FC/DC...(II)

(I) AND (II) taken together prove the result..