Question

In the fig DF II BO ,EF II CO Prove that
DE II BC​

Answer 1

Answer:

Given: DF║BO AND EF║CO

To prove:DE║BC

Proof: In ΔABO, DF║BO

           AD/BD=AF/OF ..... eq 1  (BPT)

            In ΔACO, EF║CO

             AE/EC=AF/OF  .....eq 2 (BPT)

           from eq 1 and eq 2,

            AD/BD=AE/EC

            ⇒DE║BC            (converse of BPT)

             

Step-by-step explanation: