题目内容
如图,在△ABC中,∠A=90°.
(1)利用直尺和圆规,作线段的垂直平分线,分别交BC、AB于点D、E;(保留作图痕迹,不写作法)
(2)根据(1)中所画图形,求证:BE2=AC2+AE2.

(1)利用直尺和圆规,作线段的垂直平分线,分别交BC、AB于点D、E;(保留作图痕迹,不写作法)
(2)根据(1)中所画图形,求证:BE2=AC2+AE2.

(1)如图所示:直线DE即为所求作的图形;

(2)连接CE,
∵DE是BC的垂直平分线,
∴BE=EC,
∵∠A=90°,
∴在Rt△ACE中,BE2=CE2=AC2+AE2.


(2)连接CE,
∵DE是BC的垂直平分线,
∴BE=EC,
∵∠A=90°,
∴在Rt△ACE中,BE2=CE2=AC2+AE2.

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