ABCD rectangle angle BAC = 35 ° Find the angles between the diagonals.

Consider the BAC triangle and its angles.
<BAC = 35 ° by condition.

The triangle BAC is isosceles, since BO = AO, since BO and AO are halves of the diagonals of the ABCD rectangle according to the sign that the diagonals in the rectangle are divided by the point of intersection O in half.

Let’s use the property of the rectangle diagonals: the rectangle diagonals are equal. That is, AC = BD, and BO = AO.
It follows from this that the angles at the base of the isosceles triangle ABO are equal:

<DBA = <BAC = 35 °.
From the ABO triangle, we find the angle between the diagonals of the rectangle:

<BOA = 180 ° – <DBA – <BAC = 180 ° – 35 ° – 35 ° = 110 °.

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