In an isosceles triangle ABC with base AC, the height BK meets the median AM at point O

In an isosceles triangle ABC with base AC, the height BK meets the median AM at point O. Subtract the lengths of the segments BO and OK if AB = 5cm, AC = 6cm.

The height BK of an isosceles triangle is also its median, then AK = CK = AC / 2 = 6/2 = 3 cm, and triangles ABK and CBK are rectangular.

Then, by the Pythagorean theorem, BK ^ 2 = AB ^ 2 – AK ^ 2 = 25 – 9 = 16.

BK = 4 cm.

By the property of the medians, the point O of their intersection, we divide the medians in the ratio of 2/1.

Then BО = 2 * OK.

BО + OK = 4 cm.

2 * OK + OK = 4 cm.

3 * OK = 4 cm.

OK = 4/3 = 1 (1/3) cm.

BО = 4 – 4/3 = 8/3 = 2 (2/3) cm.

Answer: The length of the segment OK = 1 (1/3) cm, the length of BО = 2 (2/3) cm.