# In parallelogram ABCD, angle B is obtuse, the parallelogram perimeter is 22. The heights BH and BK divide angle

In parallelogram ABCD, angle B is obtuse, the parallelogram perimeter is 22. The heights BH and BK divide angle B into three equal parts. find the perimeter of the quadrilateral BHDK.

1. Angle HBK = angle CBK = 90 °: 2 = 45 °. The ABH angle is also 45 °.

2. Angle BAH = 180 ° – 45 ° – 90 ° = 45 °.

3. Angle BCK = 180 ° – 45 ° – 90 ° = 45 °.

4. Triangles ABH and BCK are isosceles, since the angles at their bases are equal.

Therefore, AH = BH and BC = BK.

6. DK = CD – CK = CD – BK.

7. Perimeter BKDH = DH + DK + BH + BK. We substitute here the values of DН and DК:

Perimeter BKDH = AD – BH + CD – BK + BH + BK = AD + CD.

8.AD + CD = 22: 2 = 11 cm.

9. The perimeter of BHDK is 11 cm.

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