# The height of the parallelogram divides the angle by two angles, the difference is 20 find the angles of the parallelogram.

Given: ABCD – parallelogram;

BH – height;

<HBC – <ABH = 20 °;

Find: <A, <B, <C, <D.

By the property of the angles of a parallelogram, its opposite angles are pairwise equal, i.e. <A = <C, <B = <D.

Height BH divides angle B into two angles <HBC and <ABH, the difference between which is 20 ° by condition.

The height BH drawn to side AD forms an angle <BHD = 90 °. If parallel to draw the height DH1 from the vertex D, equal to BH, then at the intersection with BC it also forms a right angle DH1B.

Thus, two heights drawn to opposite sides of the parallelogram form a rectangle HBH1D. All angles in it are straight, namely:

<HBH1 = <H1DH = 90 °.

Hence, <ABH = 90 ° – 20 ° = 70 °.

Then the angle <B = <HBC + <ABH = 90 ° + 70 ° = 160 °. Those. <B = <D = 160 °.

Because a parallelogram is a quadrangle, then the sum of all its angles is 360 °.

Therefore, <A + <C = 360 ° – <B + <D = 360 ° – 160 ° * 2 = 360 ° – 320 ° = 40 °. And since <A = <C, then <A = <C = 40 °: 2 = 20 °.

Answer: <A = <C = 20 °, <B = <D = 160 °.