In parallelogram ABCD, the heights drawn from vertex B are equal. Prove that the given parallelogram is a rhombus.

If ABCD is a parallelogram, then its opposite angles and sides are equal. By the condition of the height BH = BK.

In triangles ABH and BCK, the angle BAN = BCК, then the angle ABH = CBH, and then the triangles ABH and BCK are equal in the legs BH and BK and the angles ABH and СBK adjacent to them, according to the second sign of equality of right-angled triangles, and then AB = BC.

And since AB = CD, BC = AD, then AB = BC = CD = AD, and therefore ABCD is a rhombus, which was required to be proved.