Question

applying the side - splitter theorem
what is the length of \\(\overline{gd}\\)?
\\(gd = \square\\)

Explanation:

Step1: Identify the Side - Splitter Theorem

The Side - Splitter Theorem states that if a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. In this triangle, since \(EH\parallel FG\) (the marks indicate parallel lines), we have \(\frac{EG}{GD}=\frac{FH}{HD}\)? Wait, no, actually, looking at the segments: \(EF\) and \(GH\) are parallel, so the line \(GH\) divides \(ED\) and \(FD\) proportionally. Wait, the lengths: \(EG = 6\), \(FH=6\), \(HD = 14\). Wait, actually, the segments on \(EF\): \(FH = 6\), and the segments on \(ED\): \(EG=6\). Wait, maybe the correct proportion is \(\frac{EG}{GD}=\frac{FH}{HD}\)? No, wait, let's re - examine. The Side - Splitter Theorem: If a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides into segments of proportional length. So in triangle \(EFD\), line \(GH\) is parallel to \(EF\) (the parallel marks). So \(GH\parallel EF\), so \(\frac{EG}{GD}=\frac{FH}{HD}\)? Wait, no, \(FH = 6\), \(HD=14\), \(EG = 6\). Wait, that can't be. Wait, maybe the segments are \(EG = 6\), \(GD=x\), \(FH = 6\), \(HD = 14\). Wait, no, the correct proportion is \(\frac{EG}{GD}=\frac{FH}{HD}\)? Wait, no, actually, the two sides are \(ED\) (with \(EG\) and \(GD\)) and \(FD\) (with \(FH\) and \(HD\)). Since \(GH\parallel EF\), by the Side - Splitter Theorem, \(\frac{EG}{GD}=\frac{FH}{HD}\). Wait, but \(FH = 6\), \(HD = 14\), \(EG = 6\). Wait, that would give \(\frac{6}{x}=\frac{6}{14}\), which would imply \(x = 14\), but that seems off. Wait, maybe I got the segments wrong. Wait, the length from \(E\) to \(G\) is 6, from \(F\) to \(H\) is 6, and from \(H\) to \(D\) is 14. So the line \(GH\) is parallel to \(EF\), so triangle \(EGH\) (no, wait, triangle \(EFD\) with \(GH\) parallel to \(EF\)). So the ratio of \(EG\) to \(GD\) should be equal to the ratio of \(FH\) to \(HD\). Wait, \(EG = 6\), \(FH = 6\), so the ratio is \(6:6 = 1:1\), so \(GD=HD = 14\)? Wait, but that would make \(ED=EG + GD=6 + 14 = 20\), and \(FD=FH+HD = 6 + 14 = 20\), so triangle \(EFD\) is isoceles? Maybe. Alternatively, maybe the segments are \(EG = 6\), \(GD=x\), \(FH = 6\), \(HD = 14\), and since \(GH\parallel EF\), \(\frac{EG}{GD}=\frac{FH}{HD}\), so \(\frac{6}{x}=\frac{6}{14}\), so \(x = 14\).

Step2: Apply the Side - Splitter Theorem

The Side - Splitter Theorem formula is \(\frac{EG}{GD}=\frac{FH}{HD}\). We know \(EG = 6\), \(FH = 6\), and \(HD = 14\). Substituting the values into the formula: \(\frac{6}{GD}=\frac{6}{14}\). Cross - multiplying gives \(6\times14=6\times GD\). Dividing both sides by 6, we get \(GD = 14\).

Answer:

\(14\)