1. **Stating the problem:** We are given a right triangle ABC with angles $A=50^\circ$, $C=40^\circ$, and side lengths $AB=8$, $BC=9$, and $BF=1$. A perpendicular from $H$ to $BC$ at $G$ creates right angles at $B$, $F$, and $G$. Segment $DE$ is parallel to $BC$ with length 6. We need to find the lengths of segments $AH$, $HG$, $AC$, $EF$, and others marked with question marks. 2. **Relevant formulas and rules:** - Use trigonometric ratios: $\sin(\theta) = \frac{o}{h}$, $\cos(\theta) = \frac{a}{h}$, $\tan(\theta) = \frac{o}{a}$ where $o$ is opposite side, $a$ is adjacent side, and $h$ is hypotenuse. - Use properties of parallel lines and similar triangles. - Use Pythagorean theorem: $a^2 + b^2 = c^2$ for right triangles. 3. **Find side $AC$:** - Triangle $ABC$ has sides $AB=8$, $BC=9$, and angle $A=50^\circ$. - Use Law of Cosines or trigonometry to find $AC$. - Since $AB$ and $BC$ are given, and angles $A$ and $C$ sum to $90^\circ$, triangle is right angled at $B$. - Check if $AB^2 + BC^2 = AC^2$: $$8^2 + 9^2 = 64 + 81 = 145$$ - So, $AC = \sqrt{145} \approx 12.04$. 4. **Find $AH$ and $HG$:** - $H$ is a point on $AC$ such that $HG$ is perpendicular to $BC$ at $G$. - Since $DE$ is parallel to $BC$, triangles involving $DE$ and $BC$ are similar. - Use similarity ratios to find $AH$ and $HG$. 5. **Find $EF$:** - Since $DE$ is parallel to $BC$ and $DE=6$, use similarity ratios between triangles to find $EF$. 6. **Summary:** - $AC = \sqrt{145} \approx 12.04$ - Use trigonometric ratios and similarity to find other segments. **Final answers:** - $AC \approx 12.04$ - $AH$, $HG$, $EF$ require additional information or calculations based on the figure and similarity. Since the problem is complex and involves multiple unknowns, the key step is to use trigonometry and similarity to find each segment step-by-step.