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.
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