In an isosceles triangle ABC AB = BC = 10, AC = 16. Find the distance between the point

In an isosceles triangle ABC AB = BC = 10, AC = 16. Find the distance between the point of intersection of the medians and the point of intersection of the bisectors of the triangle.

1. Construction: CF – median, CK – bisector of angle C, BH – median and bisector simultaneously (since triangle ABC is isosceles by the condition), M – intersection point of medians, K – intersection point of bisectors.
2. Consider a right-angled triangle BHC (angle H is a straight line).
Find the side НС = AC / 2 = 16/2 = 8.
By the Pythagorean theorem, we calculate the leg BH = (BC ^ 2 – HC ^ 2) ^ (1/2) = (100 – 64) ^ (1/2) = 6.
3. Let’s use the following property: the medians of a triangle intersect at one point, which divides each of them into two parts in a ratio of 2: 1, counting from the vertex. That is BM = 2 * MH.
Let MH = x, BM = 2 * x, then BH = BM + MH = 3 * x = 6. Hence x = 2 = MH, BM = 4.
4. The bisector divides the opposite side into parts proportional to the adjacent sides:
BK / KH = BC / CN = 10/8.
KH = BH – BK = 6 – BK. Then BK / KH = BK / (6 – BK) = 10/8.
8 * BK = 10 * (6 – BK);
18 * BK = 60; BK = 10/3.
5. BM = 4, BK = 10/3. Then MK = 4 – 10/3 = 2/3.
Answer: the distance between the point of intersection of the medians and the point of intersection of the bisectors is 2/3.