question_answer

In the figure if \[DE\parallel OB\] and \[EF\parallel BC\], then prove that \[DF\parallel OC\]

Answer:

Given, In \[\Delta \,ABC,\text{ }DE\parallel OB\] and \[EF\parallel BC\]

To Prove: \[DF\parallel OC\]

Proof: In \[\Delta \,AOB\],

\[DE\parallel OB\]

\[\therefore \frac{AE}{EB}=\frac{AD}{DO}\]                   ?(i) [Thales? Theorem]

Similarly, in \[\Delta \,ABC,\]

\[EF\parallel BC\]

\[\therefore \frac{AE}{EB}=\frac{AF}{FC}\]                    ?(ii) [Thales? Theorem]

From (i) and (ii),

\[\frac{AD}{DO}=\frac{AF}{FC}\]

\[\therefore DF\parallel OC\]                     [By Converse of Thales? Theorem] Hence Proved.