In the given figure, \(DE \parallel AB\) and \(DF \parallel AC\). Prove that \(EF \parallel BC\).
Proof:
In \(\triangle AGB\), \(DE \parallel AB\), then by Thales theorem We have,
\(\frac{AD}{DG} =\) ---- (\(1\))
In \(\triangle AGC\), \(DF \parallel AC\), then by Thales theorem, we have:
\(\frac{AD}{DG} =\) ---- (\(2\))
From equations (\(1\)) and (\(2\)), we have:
\(\frac{BE}{EG} =\)
Applying converse of Thales theorem We have,
\(EF \parallel\)
Hence, we proved.
\(\frac{AD}{DG} =\) ---- (\(1\))
In \(\triangle AGC\), \(DF \parallel AC\), then by Thales theorem, we have:
\(\frac{AD}{DG} =\) ---- (\(2\))
From equations (\(1\)) and (\(2\)), we have:
\(\frac{BE}{EG} =\)
Applying converse of Thales theorem We have,
\(EF \parallel\)
Hence, we proved.
Answer variants:
\(BC\)
\(\frac{CF}{FG}\)
\(\frac{BE}{EG}\)
\(\frac{CF}{FG}\)