In parallelogram ABCD, the bisectors of angles A and B intersect sides BC and AD at points

| September 20, 2021 | education

In parallelogram ABCD, the bisectors of angles A and B intersect sides BC and AD at points E and F, respectively. Prove that ABCD is a rhombus.

Consider the resulting parallelogram ABEF, its diagonals AE and BF intersect at right angles <BOA = <AOF.

Let us prove this, O is the intersection point of the diagonals AE and BF. <AOB = 90 ° since the angles of the triangle are ABO; (<ABO + <BAO) = (<ABE + <BAF) / 2 = 180 ° / 2 = 90 °. Here we have used parallel lines AD and BC, which add up to 180 °.

BO = OF, since in triangle ABF the height of AO is the bisector (by condition), and the median, that is, BO = OF, the diagonals are halved.

This means that if the diagonals of a parallelogram intersect at an angle of 90, then this parallelogram is a rhombus.



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