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(N/A) Given: $ABCD$ is a cyclic quadrilateral where diagonals $AC$ and $BD$ are diameters of the circle.Since $AC$ and $BD$ are diameters,they pass th

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(N/A) Given: $ABCD$ is a cyclic quadrilateral where diagonals $AC$ and $BD$ are diameters of the circle.
Since $AC$ and $BD$ are diameters,they pass through the center $O$ and are equal in length $(AC = BD)$.
We know that an angle in a semicircle is a right angle.
Since $AC$ is a diameter,$\angle ABC = 90^{\circ}$ and $\angle ADC = 90^{\circ}$.
Since $BD$ is a diameter,$\angle BAD = 90^{\circ}$ and $\angle BCD = 90^{\circ}$.
Thus,all interior angles of the quadrilateral $ABCD$ are $90^{\circ}$.
$A$ quadrilateral with all angles equal to $90^{\circ}$ is a rectangle.
Therefore,$ABCD$ is a rectangle.

If the diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral,prove that it is a rectangle.

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