In an isosceles trapezoid, an acute angle of 45 degrees. The perpendicular dropped from the apex

In an isosceles trapezoid, an acute angle of 45 degrees. The perpendicular dropped from the apex of the obtuse angle to the larger base has a length of 8 cm, and the middle line is 12 cm, find the base of the trapezoid

Let’s draw the heights BK and СK of the trapezoid. Since the trapezoid is isosceles, then the angle CDA = BCA = 45, then the right-angled triangles ABH and CDP are isosceles and AH = BH = 8 cm, DР = СР = 8 cm.Since BCРH is a rectangle, then HP = BC, then AD = (AН + BC + DP) = (BC + 16).

Then the middle line of the trapezoid is: KM = (BC + BC + 18) / 2 = 12 cm.

2 * BC = 24 – 16 = 8 cm.

BC = 8/2 = 4 cm.

BP = 8 + 4 + 8 = 20 cm.

Answer: The bases of the trapezoid are 4 cm and 20 cm.