In rhombus ABCD, the height BK is drawn from the vertex of the obtuse angle B to the side AD

| May 1, 2021 | education

In rhombus ABCD, the height BK is drawn from the vertex of the obtuse angle B to the side AD and the height BP is drawn to the side CD. Prove the equality of triangles ABK and CBP, and equality of triangles KBP and BAD

Since BK and BP are heights, the triangles ABK and BСР are rectangular.

In a rhombus, all sides are equal, then AB = BC.

The opposite angles of the rhombus are equal, then the angle ABK = HРВ.

Then right-angled triangles ABK and BCP along the hypotenuse and acute angles.

The KBP triangle cannot be equal to the BAD triangle, since BK is the leg of the right-angled triangle ABK, in which AB is the hypotenuse.

The KBP triangle is isosceles, since BK = BP as the height of the rhombus. Let us prove that the triangle KBP and ABD are similar. Since AB = AD, and BK = BP, the sides of the triangles are proportional.

Let the angle BAD = X, then the angle ADS = 180 – X. In the quadrangle BKDP, the angle KВP = 360 – 90 – 90 – (180 – X) = X. Then the angle BAD = KBP, which means that the triangle BAD is similar to the triangle KBP in two proportional to the sides and the angle between them. Q.E.D.



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