AB = 52, P_ABC = ?

The image shows a right-angled triangle ABC, with a circle inscribed in it. The radius of the inscribed circle is 8. The side AB is given as 52. The question asks for the perimeter of triangle ABC. Let the points of tangency of the inscribed circle with sides BC, AC, and AB be D, E, and F respectively. Let O be the center of the inscribed circle. Since CK is perpendicular to AC and CK is the radius, CK = 8. Similarly, let BD = BF and AE = AF. Also, CD = CE and the radius to the point of tangency is perpendicular to the tangent. Thus, OCKD is a square with side length 8. Therefore, CD = CE = 8. Since the triangle is right-angled at C, AC = AE + EC and BC = BD + DC. The perimeter P_ABC = AB + BC + AC = AB + (BF + FD) + (AE + EC). Since FD is a diameter of the circle and is parallel to AC, FD = 2 * radius = 16. However, this is incorrect as FD is not a diameter. Let's use the property that tangents from a vertex to an inscribed circle are equal in length. Let AF = x, BF = y, CE = z. Then AC = x + z, BC = y + z, AB = x + y. The radius of the inscribed circle in a right-angled triangle is given by r = (a + b - c) / 2, where a and b are the legs and c is the hypotenuse. In our case, let AC = b, BC = a, AB = c. So, 8 = (a + b - 52) / 2. This gives a + b = 16 + 52 = 68. The perimeter P_ABC = a + b + c = 68 + 52 = 120.