如图10,在△ABC中,AB=2,∠BDC=90°,BD=CD,∠ABD=∠ACD=15°,AD平分∠BAC.
(1)试猜想△ABC的形状,并说明理由(2)求BD的长.(3)以BC所在的直线为x轴,AD所在的直线为y轴,建立平面直角坐标系,写出图中各个点的坐标.
人气:288 ℃ 时间:2020-03-25 05:15:35
解答
1、△ABC是等边三角形证明:∵BD=CD,∠BDC=90∴∠DBC=∠DCB=45∵∠ABD=∠ACD=15∴∠ABC=∠ABD+∠DBC=15+45=60∠ACB=∠ACD+∠DCB=15+45=60∴等边△ABC延长AD交BC于E∵等边△ABC∴BC=AB∵AB=2∴BC=2∵A...
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