In rectangle ADCD with side AD = 9 cm, the bisector AF cuts off the segment BF = 5 cm

In rectangle ADCD with side AD = 9 cm, the bisector AF cuts off the segment BF = 5 cm from side BC. Find the perimeter of the rectangle.

Angle A is straight and is equal to 90 °. The bisector AF divides angle A in half, so:

angle BAC = 90/2 = 45 °;

angle BFA = 180 ° – angle ABC – angle BAF = 180 – 90 – 45 = 45 °;

angle BAF = angle BFA = 45 °, therefore triangle ABF is isosceles;

AB – BF = 5 cm;

The perimeter of the rectangle is P = AB * AD = 5 * 9 = 45 cm.