# In rectangle ABCD, side AB is 6 cm, diagonal AC is 10 cm. O is the intersection point of the diagonals.

**In rectangle ABCD, side AB is 6 cm, diagonal AC is 10 cm. O is the intersection point of the diagonals. The perpendicular BH is dropped on the diagonal. Find the segments into which the diagonal AC of points H and O are divided.**

Since AВСD is a rectangle, its diagonals, at point O, are divided in half. Then AO = CO = AC / 2 = 10/2 = 5 cm.Since AC = ВD, then BO = AO = 5 cm.

The height of the ВН forms two right-angled triangles, AВН and ВOН, in which, according to the Pythagorean theorem, we express the ВН. Let the length AH = X cm, then OH = (5 – X) cm.

BH ^ 2 = AB ^ 2 – AH ^ 2 = 36 – X2. (1).

BH ^ 2 = BO ^ 2 – OH ^ 2 = 25 – (5 – X) 2. (2).

Equate Equations 1 and 2.

36 – X ^ 2 = 25 – (5 – X) ^ 2.

36 – X ^ 2 = 25 – 25 + 10 * X – X62.

10 * X = 36.

X = AH = 36/10 = 3.6 cm.

OH = 5 – 3.6 = 1.4 cm.

CH = OH + OС = 1.4 + 5 = 6.4 cm.

Answer: Point O divides the diagonal into segments 5 cm and 5 cm, point H divides the diagonal and segments 3.6 cm and 6.4 cm.