In the parallelogram ABCD, the heights BH and BE are drawn to the sides AD and CD

| August 14, 2021 | education

In the parallelogram ABCD, the heights BH and BE are drawn to the sides AD and CD, respectively, and BH is equal to BE, prove that ABCD is a rhombus.

Since BH and ve are the height of the parallelogram, the triangles ABH and BCE are rectangular.

Let us prove that triangle ABH is equal to triangle BCE. Since the parallelogram has opposite angles, the angle BAН = BCE, and since the angle BСE = AHB = 90, then the angle ABH = CBE = (90 – BAН).

By condition, BH = BE, then triangles ABH and BCE are equal in leg and acute angle, the second sign of equality of right-angled triangles. Then AB = BC.

Since the opposite sides of the parallelogram are equal, then AB = BC = CD = AD.

If all sides of a parallelogram are equal, then such a parallelogram is a rhombus, as required.



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