The bisector LB and the height LH are drawn in the parallelogram KLMN. Base B of bisector LB

The bisector LB and the height LH are drawn in the parallelogram KLMN. Base B of bisector LB and base H of height LH lie on side KN. The larger angle of the parallelogram is 126 degrees. Find the angle between LB and LH.

Since, by condition, LB is the bisector of the angle KLM, then the angle BLM = KLM / 2 = 126/2 = 63.

The angle LBK is equal to the angle BLM as the intersecting angles at the intersection of the parallel lines LM and KN of the secant LB.

The LHB triangle is rectangular, since LH is the height of the trapezoid, then the angle BLH = 180 – LHB – LBH = 180 – 90 – 63 = 27.

Answer: The angle between LB and LH is 27.