Question:

Step 1: Identify the triangle dimensions. Let the sides of the triangle be $a=6$, $b=7$, and $c=8$. The shortest side is $a=6$. Step 2: Calculate the semi-perimeter. The semi-perimeter $s$ is given by: $$s = \frac{a+b+c}{2} = \frac{6+7+8}{2} = \frac{21}{2}$$ Step 3: Calculate the area of the triangle. Using Heron's formula, the area $A$ is: $$A = \sqrt{s(s-a)(s-b)(s-c)}$$ First, calculate the terms $(s-a)$, $(s-b)$, and $(s-c)$: $$s-a = \frac{21}{2} - 6 = \frac{21-12}{2} = \frac{9}{2}$$ $$s-b = \frac{21}{2} - 7 = \frac{21-14}{2} = \frac{7}{2}$$ $$s-c = \frac{21}{2} - 8 = \frac{21-16}{2} = \frac{5}{2}$$ Now, substitute these values into Heron's formula: $$A = \sqrt{\frac{21}{2} \cdot \frac{9}{2} \cdot \frac{7}{2} \cdot \frac{5}{2}} = \sqrt{\frac{3 \cdot 7 \cdot 3^2 \cdot 7 \cdot 5}{16}} = \sqrt{\frac{3^3 \cdot 5 \cdot 7^2}{16}} = \frac{3 \cdot 7 \sqrt{3 \cdot 5}}{4} = \frac{21\sqrt{15}}{4}$$ Step 4: Calculate the inradius. The inradius $r$ is given by the formula $r = A/s$: $$r = \frac{21\sqrt{15}/4}{21/2} = \frac{21\sqrt{15}}{4} \cdot \frac{2}{21} = \frac{\sqrt{15}}{2}$$ Step 5: Calculate the altitude to the shortest side. The altitude $h_a$ from vertex $A$ to side $a$ is given by $h_a = 2A/a$: $$h_a = \frac{2(21\sqrt{15}/4)}{6} = \frac{21\sqrt{15}/2}{6} = \frac{21\sqrt{15}}{12} = \frac{7\sqrt{15}}{4}$$ Step 6: Determine the height of the similar triangle. The line segment $PQ$ is parallel to the shortest side $a$ (side $BC$) and passes through the incenter $I$. The incenter $I$ is at a perpendicular distance $r$ from side $a$. The triangle formed by vertex $A$ and the segment $PQ$ (let's call it $\triangle APQ$) is similar to the original triangle $\triangle ABC$. The altitude of $\triangle APQ$ from vertex $A$ to the line $PQ$ is the total altitude $h_a$ minus the inradius $r$: $$h_{APQ} = h_a - r = \frac{7\sqrt{15}}{4} - \frac{\sqrt{15}}{2} = \frac{7\sqrt{15}}{4} - \frac{2\sqrt{15}}{4} = \frac{5\sqrt{15}}{4}$$ Step 7: Calculate the ratio of similarity. The ratio of similarity $k$ between $\triangle APQ$ and $\triangle ABC$ is the ratio of their corresponding altitudes: $$k = \frac{h_{APQ}}{h_a} = \frac{5\sqrt{15}/4}{7\sqrt{15}/4} = \frac{5}{7}$$ Step 8: Calculate the length of the segment PQ. Since $\triangle APQ$ is similar to $\triangle ABC$ with ratio $k$, the length of $PQ$ is $k$ times the length of the corresponding side $a$: $$PQ = k \times a = \frac{5}{7} \times 6 = \frac{30}{7}$$