The diagonal of an isosceles trapezoid divides its acute angle in half. the base of the trapezoid is 6 and 15 cm.

The diagonal of an isosceles trapezoid divides its acute angle in half. the base of the trapezoid is 6 and 15 cm. Find the perimeter of the trapezoid, the middle line.

By condition, the diagonal AC is the bisector of angle A, hence the angle BAC = CAD.

Angle CAD = BCA as criss-crossing angles at the intersection of parallel straight lines AD and BC secant AC, then angle BAC = BCD, and triangle ABC is isosceles and AB = BC = 6 cm.

Then CD = AB = BC = 6 cm.

Determine the perimeter of the trapezoid.

P = AB + BC + CD + AD = 6 + 6 + 6 + 15 = 33 cm.

Determine the length of the middle line of the MK.

MK = (AD + BC) / 2 = (15 + 6) / 2 = 21/2 = 10.5 cm.

Answer: The perimeter of the trapezoid is 33 cm, the middle line is 10.5 cm.