Question

$\overline{ij} \parallel \overline{hk}$. find $gj$.

Explanation:

Step1: Identify Similar Triangles

Since \(\overline{IH} \parallel \overline{HK}\) (wait, actually \(\overline{IH}\) and \(\overline{HK}\) – no, the given is \(\overline{ID} \parallel \overline{HK}\)? Wait, the diagram has \(\overline{ID} \parallel \overline{HK}\)? Wait, the problem says \(\overline{ID} \parallel \overline{HK}\)? Wait, the original problem: \(\overline{ID} \parallel \overline{HK}\)? Wait, the diagram: the two parallel lines are the ones with the ticks, so the large triangle and the smaller triangle are similar by the Basic Proportionality Theorem (Thales' theorem) because \(HK \parallel ID\). So triangles \(GHK\) and \(GID\) are similar.

Step2: Set Up Proportion

Let \(GJ = x\), so \(GK = 15\), and \(GJ = GK + KJ\)? Wait, no, wait: \(GJ\) is the side, \(GK = 15\), and the sides of the similar triangles: the ratio of corresponding sides. The sides \(GH = 44\), \(GI = 77\), \(GK = 15\), \(GJ = x\). So by similarity, \(\frac{GH}{GI} = \frac{GK}{GJ}\). Wait, no: \(\frac{GH}{GI} = \frac{GK}{GJ}\)? Wait, \(GH = 44\), \(GI = 77\), \(GK = 15\), \(GJ = x\). So \(\frac{44}{77} = \frac{15}{x}\)? Wait, no, that can't be. Wait, maybe \(\frac{GH}{GI - GH} = \frac{GK}{KJ}\)? Wait, no, let's re-examine.

Wait, the segments: \(GI = 77\), \(GH = 44\), so \(HI = GI - GH = 77 - 44 = 33\). The line \(HK\) is parallel to \(ID\), so by the Basic Proportionality Theorem (Thales' theorem), \(\frac{GH}{HI} = \frac{GK}{KJ}\). Wait, \(GK = 15\), \(KJ = GJ - GK = x - 15\). So \(\frac{44}{33} = \frac{15}{x - 15}\). Wait, \(\frac{44}{33} = \frac{4}{3}\), so \(\frac{4}{3} = \frac{15}{x - 15}\). Cross-multiplying: \(4(x - 15) = 45\) → \(4x - 60 = 45\) → \(4x = 105\) → \(x = 26.25\)? No, that's wrong. Wait, maybe the ratio is \(\frac{GH}{GI} = \frac{GK}{GJ}\). So \(GH = 44\), \(GI = 77\), \(GK = 15\), \(GJ = x\). So \(\frac{44}{77} = \frac{15}{x}\)? Wait, \(\frac{44}{77} = \frac{4}{7}\), so \(\frac{4}{7} = \frac{15}{x}\) → \(4x = 105\) → \(x = 26.25\)? No, that doesn't make sense. Wait, maybe I got the segments wrong. Wait, \(GJ\) is the entire side, \(GK = 15\), and the sides \(GH = 44\), \(GI = 77\). So the ratio of similarity is \(\frac{GH}{GI} = \frac{44}{77} = \frac{4}{7}\). So the ratio of \(GK\) to \(GJ\) should be \(\frac{4}{7}\). So \(\frac{15}{GJ} = \frac{4}{7}\)? No, that would be if \(GK\) corresponds to \(GH\), but no. Wait, maybe the triangles are \(GHK\) and \(GIJ\)? Wait, no, the parallel lines are \(HK \parallel ID\), so the triangles are \(GHK\) and \(GID\). So \(GH\) corresponds to \(GI\), \(HK\) corresponds to \(ID\), and \(GK\) corresponds to \(GJ\). So \(\frac{GH}{GI} = \frac{GK}{GJ}\). So \(GH = 44\), \(GI = 77\), \(GK = 15\), \(GJ = x\). So \(\frac{44}{77} = \frac{15}{x}\). Simplify \(\frac{44}{77} = \frac{4}{7}\). So \(\frac{4}{7} = \frac{15}{x}\) → \(4x = 105\) → \(x = 26.25\)? That can't be. Wait, maybe I mixed up the segments. Wait, \(GI = 77\), \(GH = 44\), so \(HI = 77 - 44 = 33\). Then, by Thales' theorem, \(\frac{GH}{HI} = \frac{GK}{KJ}\). So \(GH = 44\), \(HI = 33\), \(GK = 15\), \(KJ = x - 15\) (since \(GJ = GK + KJ = 15 + (x - 15) = x\)). So \(\frac{44}{33} = \frac{15}{x - 15}\). Simplify \(\frac{44}{33} = \frac{4}{3}\). So \(\frac{4}{3} = \frac{15}{x - 15}\) → \(4(x - 15) = 45\) → \(4x - 60 = 45\) → \(4x = 105\) → \(x = 26.25\). Wait, that's 105/4 = 26.25. But that seems odd. Wait, maybe the ratio is \(\frac{GI}{GH} = \frac{GJ}{GK}\). So \(\frac{77}{44} = \frac{GJ}{15}\). Simplify \(\frac{77}{44} = \frac{7}{4}\). So \(\frac{7}{4} = \frac{GJ}{15}\) → \(GJ = \frac{7 \times 15}{4} = \frac{105}{4} = 26.25\). Wait, but 26.25 is 105/4. But maybe I made a mistake. Wait, let's check again. The two parallel lines: the transversal is \(GJ\) and \(GI\). So by the Basic Proportionality Theorem, if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. So in triangle \(GIJ\), line \(HK\) is parallel to \(IJ\) (wait, no, the parallel lines are the ones with the ticks, so \(HK \parallel ID\), so \(HK\) is parallel to \(ID\), so in triangle \(GID\), \(HK\) is parallel to \(ID\), so \(HK\) intersects \(GI\) at \(H\) and \(GJ\) at \(K\). Therefore, \(\frac{GH}{GI} = \frac{GK}{GJ}\). So \(GH = 44\), \(GI = 77\), \(GK = 15\), \(GJ = x\). So \(\frac{44}{77} = \frac{15}{x}\). Simplify \(\frac{44}{77} = \frac{4}{7}\). So \(\frac{4}{7} = \frac{15}{x}\) → \(x = \frac{15 \times 7}{4} = \frac{105}{4} = 26.25\). Wait, but 26.25 is 105/4. Alternatively, maybe the ratio is \(\frac{GI - GH}{GH} = \frac{GJ - GK}{GK}\). So \(\frac{77 - 44}{44} = \frac{x - 15}{15}\). So \(\frac{33}{44} = \frac{x - 15}{15}\). Simplify \(\frac{3}{4} = \frac{x - 15}{15}\) → \(3 \times 15 = 4(x - 15)\) → \(45 = 4x - 60\) → \(4x = 105\) → \(x = 26.25\). So that's the same result.

Answer:

\(26.25\) (or \(\frac{105}{4}\))