The height lowered from the top of the obtuse angle of the parallelogram equal to 135 degrees is 4 cm

The height lowered from the top of the obtuse angle of the parallelogram equal to 135 degrees is 4 cm, and divides the side to which it is lowered into two equal parts. Find this side.

ABCD is a parallelogram. ∠B = ∠D = 135 °. BH = 4 cm – lowered height to the AD side.
1. Since the opposite angles of the parallelogram are equal, we denote ∠A and ∠C as x. By the theorem on the sum of the angles of a quadrangle:
∠A + ∠B + ∠C + ∠D = 360 °;
x + 135 ° + x + 135 ° = 360 °;
2 * x = 360 ° – 270 °;
2 * x = 90 °;
x = 90 ° / 2;
x = 45 °.
Thus, ∠A = ∠C = x = 45 °.
2. Consider △ AHB: ∠AHB = 90 ° (since BH is height), ∠HAB (aka ∠A) = 45 °.
By the theorem on the sum of the angles of a triangle:
∠AHB + ∠HAB + ∠ABH = 180 °;
90 ° + 45 ° + ∠ABH = 180 °;
∠ABH = 180 ° – 135 °;
∠ABH = 45 °.
Since in △ AHB ∠HAB = ∠ABH = 45 °, △ AHB is isosceles, and ∠HAB and ∠ABH are the angles at the base of AB of an isosceles triangle. Then HA = HB = 4 cm – lateral sides.
3. Point H divides the base AD in half, then:
AD = 2 * HA = 2 * 4 = 8 (cm).
Answer: 8 cm.