AC = BC, CD : DA = 3:2, P_triangle = ?

Let the points of tangency on AC, BC, and AB be E, F, and G respectively. Since AC = BC, triangle ABC is isosceles. Also, since the circle is inscribed, it is tangent to the sides. Let CD = 3x and DA = 2x. Then AC = CD + DA = 3x + 2x = 5x. Since AC = BC, BC = 5x. Let AG = BG = y. Then AB = 2y. The perimeter of triangle ABC is P = AC + BC + AB = 5x + 5x + 2y = 10x + 2y. From the properties of tangents from a vertex to an inscribed circle, we have CD = CE = 3x, DA = AG = 2x, and BF = BG = y. Thus, AC = CE + EA = 3x + EA = 5x, which means EA = 2x. Since EA = AG, this is consistent. Similarly, BC = CF + FB = 3x + FB = 5x, which means FB = 2x. Since FB = BG, this is consistent. So, AB = AG + GB = 2x + 2x = 4x. Therefore, the perimeter P = 10x + 4x = 14x. The question asks for P_triangle, which is the perimeter. The value 8 is given for the side AB. So, AB = 8. Thus, 4x = 8, which means x = 2. Then AC = 5x = 5(2) = 10 and BC = 5x = 5(2) = 10. The perimeter P = AC + BC + AB = 10 + 10 + 8 = 28.