$ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ as well as $\angle C$. Show that: diagonal $BD$ bisects $\angle B$ as well as $\angle D$.
(N/A) Given: $ABCD$ is a rectangle in which diagonal $AC$ bisects $\angle A$ and $\angle C$.
$1$. Since $ABCD$ is a rectangle,$AB \parallel DC$ and $AD \parallel BC$.
$2$. Since $AC$ bisects $\angle A$,$\angle DAC = \angle BAC$. Also,$\angle DAC = \angle BCA$ (alternate interior angles).
$3$. Therefore,$\angle BAC = \angle BCA$. In $\triangle ABC$,sides opposite to equal angles are equal,so $AB = BC$.
$4$. $A$ rectangle with adjacent sides equal is a square. Thus,$ABCD$ is a square.
$5$. In a square,diagonals bisect the opposite angles.
$6$. Therefore,diagonal $BD$ bisects $\angle B$ as well as $\angle D$.
$1$. Since $ABCD$ is a rectangle,$AB \parallel DC$ and $AD \parallel BC$.
$2$. Since $AC$ bisects $\angle A$,$\angle DAC = \angle BAC$. Also,$\angle DAC = \angle BCA$ (alternate interior angles).
$3$. Therefore,$\angle BAC = \angle BCA$. In $\triangle ABC$,sides opposite to equal angles are equal,so $AB = BC$.
$4$. $A$ rectangle with adjacent sides equal is a square. Thus,$ABCD$ is a square.
$5$. In a square,diagonals bisect the opposite angles.
$6$. Therefore,diagonal $BD$ bisects $\angle B$ as well as $\angle D$.
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