Question

find fh.
fh =
submit

Explanation:

Step1: Identify Similar Triangles

From the diagram, \( GJ \parallel HI \) (indicated by the red arrows), so \( \triangle FGJ \sim \triangle FHI \) by the Basic Proportionality Theorem (AA similarity, as corresponding angles are equal due to parallel lines).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. So, \( \frac{FJ}{FI} = \frac{FG}{FH} \). First, find \( FI \): \( FI = FJ + JI = 11 + 22 = 33 \). Let \( FH = x \), and \( FG = 50 \) (wait, no—wait, \( FG \) is part of \( FH \)? Wait, no, looking at the diagram: \( FG \) is 50? Wait, no, the segment from \( F \) to \( G \) is 50? Wait, no, the segment from \( H \) to \( G \) is 50? Wait, no, the diagram shows \( HG = 50 \)? Wait, no, let's re-express. Wait, \( FJ = 11 \), \( JI = 22 \), so \( FI = 11 + 22 = 33 \). \( FG \) and \( FH \): since \( GJ \parallel HI \), \( \triangle FGJ \sim \triangle FHI \), so \( \frac{FJ}{FI} = \frac{FG}{FH} \). Wait, \( FG \) is the length from \( F \) to \( G \), and \( FH \) is from \( F \) to \( H \). Wait, maybe \( HG = 50 \), so \( FH = FG + GH \)? No, wait, the diagram: \( H \) to \( G \) is 50? Wait, no, the segment labeled 50 is from \( H \) to \( G \)? Wait, no, the points: \( H \), \( G \), \( F \) and \( H \), \( I \), \( F \)? Wait, maybe \( G \) is on \( FH \) and \( J \) is on \( FI \), with \( GJ \parallel HI \). So \( FG \) is a part of \( FH \), and \( FJ \) is a part of \( FI \). So \( FG = FH - HG \)? No, maybe \( HG = 50 \), so \( FH = FG + GH \)? Wait, no, let's use the similarity ratio. \( FJ = 11 \), \( FI = FJ + JI = 11 + 22 = 33 \). The ratio of \( FJ \) to \( FI \) is \( \frac{11}{33} = \frac{1}{3} \). Therefore, the ratio of \( FG \) to \( FH \) should also be \( \frac{1}{3} \). Wait, but \( FG \) is \( FH - HG \)? No, wait, maybe \( FG \) is 50? Wait, no, the segment from \( F \) to \( G \) is 50? Wait, the diagram shows \( H \) to \( G \) as 50? Wait, I think I misread. Let's correct: Let \( FG = y \), \( FH = y + 50 \)? No, no, the diagram: \( H \) to \( G \) is 50, \( F \) to \( G \) is... Wait, no, the points are \( H \), \( G \), \( F \) and \( H \), \( I \), \( F \), with \( G \) on \( FH \) and \( J \) on \( FI \), and \( GJ \parallel HI \). So by the Basic Proportionality Theorem (Thales' theorem), \( \frac{FJ}{FI} = \frac{FG}{FH} \). \( FJ = 11 \), \( FI = 11 + 22 = 33 \), so \( \frac{11}{33} = \frac{1}{3} \). Therefore, \( \frac{FG}{FH} = \frac{1}{3} \), so \( FH = 3 \times FG \). But wait, the segment from \( H \) to \( G \) is 50, so \( FH = FG + GH \), so \( 3 \times FG = FG + 50 \), so \( 2 \times FG = 50 \), so \( FG = 25 \), then \( FH = 25 + 50 = 75 \)? Wait, no, that doesn't make sense. Wait, maybe \( FG = 50 \), so \( FH = 3 \times FG \)? No, \( \frac{FJ}{FI} = \frac{11}{33} = \frac{1}{3} \), so \( \frac{FG}{FH} = \frac{1}{3} \), so \( FH = 3 \times FG \). If \( FG = 50 \), then \( FH = 150 \)? Wait, no, I think I mixed up the segments. Wait, the segment labeled 50 is from \( H \) to \( G \), so \( HG = 50 \), and \( FG \) is from \( F \) to \( G \), so \( FH = FG + HG \). From similarity, \( \frac{FJ}{FI} = \frac{FG}{FH} \), so \( \frac{11}{33} = \frac{FG}{FG + 50} \). Solve for \( FG \): \( \frac{1}{3} = \frac{FG}{FG + 50} \), cross-multiply: \( FG + 50 = 3 \times FG \), \( 50 = 2 \times FG \), \( FG = 25 \). Then \( FH = FG + HG = 25 + 50 = 75 \)? Wait, no, that's not right. Wait, maybe the segment labeled 50 is \( FG \), so \( FG = 50 \), and \( HG \) is what we need? No, the problem is to find \( FH \). Wait, let's start over. \( GJ \parallel HI \), so \( \triangle FGJ \sim \triangle FHI \) (AA similarity, since \( \angle F \) is common, and \( \angle FGJ = \angle FHI \) because \( GJ \parallel HI \)). Therefore, the ratio of corresponding sides is equal: \( \frac{FJ}{FI} = \frac{FG}{FH} \). \( FJ = 11 \), \( FI = FJ + JI = 11 + 22 = 33 \). \( FG \) is the length from \( F \) to \( G \), and \( FH \) is from \( F \) to \( H \). Let \( FG = x \), then \( FH = x + 50 \) (if \( HG = 50 \))? No, maybe \( FG = 50 \), so \( FH = x \), and \( FG = 50 \), so \( \frac{11}{33} = \frac{50}{x} \), so \( \frac{1}{3} = \frac{50}{x} \), so \( x = 150 \). Ah, that makes sense! I think I misassigned the segments. So \( FG = 50 \), \( FJ = 11 \), \( FI = 33 \), so \( \frac{FJ}{FI} = \frac{11}{33} = \frac{1}{3} \), so \( \frac{FG}{FH} = \frac{1}{3} \), so \( FH = 3 \times FG = 3 \times 50 = 150 \). Yes, that's the correct approach. So the similarity ratio is \( \frac{1}{3} \), so \( FH = 3 \times FG \), and \( FG = 50 \), so \( FH = 150 \).

Answer:

\( \boxed{150} \)