Given a triangle PEF with the sides PE = 3, PF = 5, EF = 7. On the continuation of the side FP

Given a triangle PEF with the sides PE = 3, PF = 5, EF = 7. On the continuation of the side FP beyond the point P, the segment PA = 1.5 is postponed. Find the distance d between the centers of the circles around triangles EPA and EAF. In the answer, indicate the number equal to 2d.

To solve this problem, we first calculate the cosine of the angle F.

cos (F) = (5 ^ 2 + 7 ^ 2 – 3 ^ 2) / (2 * 5 * 7) = 13/14.

Length squared | AE | we will find by the ready-made formula.

AE ^ 2 = 7 ^ 2 + (6.5) ^ 2 – 2 * 7 * 6.5 * 13/14 = 27/4.

AE = 3/2 * √3.

Obviously, | AE | ^ 2 + | AP | ^ 2 = | EP | ^ 2, APE – is a right-angled triangle. The center of the circle described near the triangle APE is exactly in the middle of the hypotenuse | PE |. Around triangle EAF – in the middle of the segment | EF |.

The length connecting the centers of the circles is the midline of the PEF triangle;

d = 5/2 = 2.5.

2d = 5.