在△ABC中,AB=AC,AB的垂直平分线DE与AC所在的直线相交于点E,垂足为D,连接BE.已知AE=5,tan∠AED=34

2025-03-17 06:09:51
推荐回答(1个)
回答1:

①若∠BAC为锐角,如答图1所示:

∵AB的垂直平分线是DE,
∴AE=BE,ED⊥AB,AD=
1
2
AB,
∵AE=5,tan∠AED=
3
4

∴sin∠AED=
3
5

∴AD=AE?sin∠AED=3,
∴AB=6,
∴BE+CE=AE+CE=AC=AB=6;
②若∠BAC为钝角,如答图2所示:

同理可求得:BE+CE=16.
故答案为:6或16.

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