Question

ABCD is a trapezium with AB parallel DC a line segment EF parallel ab meets AD in and BC in fact is parallel DC why​

Answer 1

Answer:

Step-by-step explanation:

Let ABCD be the trapezium in which F is the mid point of BC.  

Since AB || CD and BC is a transversal, then

∠DCF = ∠EBF     (alternate interior angles)  ....(1)∠DCF = ∠EBF     alternate interior angles  ....1

In △DCF and △EBF∠DCF = ∠EBF         [using (1)]     CF = FB                 [as, F is mid point of BC]  ∠DFC = ∠BFE       [ vertically opposite angles]so, △DCF is congruent to △EBF  (ASA)⇒DF = FE                [CPCT]⇒F is the mid point of DE