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20. (本题7分)如图,已知在$Rt\triangle ABC$中,$AB = BC$,$\angle ABC = 90^{\circ}$.
(1)尺规作图：以点B为圆心,AC的长为半径画弧,再以点C为圆心,AB长为半径画弧,两弧在BC上方交于点D,连接AD,BD,CD(不写作法,保留作图痕迹)；
(2)求证：四边形ABCD是正方形.
(1)尺规作图：以点B为圆心,AC的长为半径画弧,再以点C为圆心,AB长为半径画弧,两弧在BC上方交于点D,连接AD,BD,CD(不写作法,保留作图痕迹)；
(2)求证：四边形ABCD是正方形.
答案：
(1)解:如图即为所求:
(2)证明:由作图可知,$DB=AC,DC=AB$,
∵$BC=CB$,
$\therefore \triangle ABC\cong \triangle DCB(SSS)$.
$\therefore \angle ABC=\angle DCB$.
$\because \angle ABC=90^{\circ}$,
$\therefore \angle DCB=90^{\circ}$.
$\therefore \angle ABC+\angle DCB=180^{\circ}$,
$\therefore AB// DC$.
$\because AB=DC$,
$\therefore$四边形$ABCD$是平行四边形.
$\because AB=BC,\angle ABC=90^{\circ}$,
$\therefore$四边形$ABCD$是正方形.
(1)解:如图即为所求:
(2)证明:由作图可知,$DB=AC,DC=AB$,
∵$BC=CB$,
$\therefore \triangle ABC\cong \triangle DCB(SSS)$.
$\therefore \angle ABC=\angle DCB$.
$\because \angle ABC=90^{\circ}$,
$\therefore \angle DCB=90^{\circ}$.
$\therefore \angle ABC+\angle DCB=180^{\circ}$,
$\therefore AB// DC$.
$\because AB=DC$,
$\therefore$四边形$ABCD$是平行四边形.
$\because AB=BC,\angle ABC=90^{\circ}$,
$\therefore$四边形$ABCD$是正方形.
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