The bisector of one of the corners of the rectangle divides its side into 42 cm and 14 cm segments

The bisector of one of the corners of the rectangle divides its side into 42 cm and 14 cm segments, starting from the vertex closest to this corner. Calculate the line segments into which the same bisector divides the diagonal of the rectangle.

Since AM is the bisector of the angle BAD, it cuts off the isosceles triangle ABM, in which AB = BM = 42 cm.

Side AD = BC since ABCD is a rectangle, then AD = BM + CM = 42 + 14 = 56 cm.

By the Pythagorean theorem, we determine the value of the diagonal BD.

BD ^ 2 = AD ^ 2 + AB ^ 2 = 3136 + 1764 = 4900.

ВD = 70 cm.

Let the length of the segment BO = X cm, then DO = (70 – X) cm.

The PTO and AOD triangles are similar in two angles, then:

BM / AD = BO / DO.

42/56 = X / (70 – X).

56 * X = 2940 – 42 * X.

98 * X = 2940.

X = BO = 2940/98 = 30 cm.

DО = 70 – 30 = 40 cm.

Answer: The bisector divides the BD diagonal into 30 cm and 40 cm segments.