In the parallelogram ABCD through the point O – the intersection of the diagonals – a straight line is drawn

In the parallelogram ABCD through the point O – the intersection of the diagonals – a straight line is drawn that intersects the sides BC and AD at points K and E, respectively BO = OE Find the angle KBE.

Angle KBE = KBO + OBE.

Consider the BOE triangle – equilateral.

Therefore,

OBE = OBE = α,

BOE = 180º – 2 * α.

The EOD triangle is isosceles, since BO = OD, and BO = OE.

Then, OED = ODE = β, EOD = 180º – 2 * β.

BOD is an extended angle and is equal to 180º, BOD = BOE + EOD.

That is, 180º = 180º – 2 * α + 180º – 2 * β, whence α + β = 90º (1).

Since BK is parallel to ED, and the corners BOK = EOD are vertical, the angle KBO = ODE = β.

Thus, KBE = KBO + OBE = β + α.

From (1) it is proved that α + β = 90º, hence KBE = 90º.