www.3112.net > 已知,在△ABC中,∠BAC=90°,∠ABC=45°,点D为直线BC上一动点(点D不与点B,C重合...

已知,在△ABC中,∠BAC=90°,∠ABC=45°,点D为直线BC上一动点(点D不与点B,C重合...

(1)证明:如图1,∵在△ABC中,∠BAC=90°,∠ABC=45°,∴∠ACB=45°,∴∠ACB=∠ABC,∴AB=AC.∵四边形ADEF为正方形,∴AD=DE=EF=AF,∠FAD=90°,∴∠BAC=∠FAD,∴∠BAC-∠DAC=∠FAD-∠DAC,∴∠BAD=∠CAF.

无论点D运动到何处,都有BC=FC+DC,理由如下:在△ABC中,∵∠BAC=90°,∠ABC=45°,∴∠ACB=45°,∴AB=AC,∵四边形ADEF是正方形,∴AD=AF,∠DAF=90°,∵∠BAD+∠DAC=∠FAC+∠DAC=90°,∴∠BAD=∠FAC,∴△BAD

参考例题:在三角形ABC中,角ABC=45°,点D(与B、C不重合)为射线BC上一动点连接AD,以AD为一边且在AD的右侧作正方形ADEF1.如果AB不等于AC,且D在线

∵∠BAC=90°,∠ABC=45°,∴∠ACB=∠ABC=45°,∴AB=AC,∵四边形ADEF是正方形,∴AD=AF,∠DAF=90°,∵∠BAD=90°∠DAC,∠CAF=90°∠DAC,∴∠BAD=∠CAF,则在△BAD和△CAF中,,∴△BAD≌△CAF(SAS),∴BD=CF,∵BD+CD

(1)证明:①∵∠BAC=90°,AB=AC,∴∠ABC=∠ACB=45°,∵四边形ADEF是正方形,∴AD=AF,∠DAF=90°,∵∠BAC=∠BAD+∠DAC=90°,∠DAF=∠CAF+∠DAC=90°,∴∠BAD=∠CAF,在△BAD和△CAF中,AB=AC∠BAD=∠CAFAD

1) 因为 角1+角3 = 角2 + 角3 = 90度,因此∠1 = ∠2. 因为 角1 = 角2,AB = AC,AD = AF,因此△ABD △ACF => BD = CF BC = BD + CD = CF + CD (2) 因为 角1 = 角2 = 90度,AB = AC,AD = AF,因此△ABD △ACF => CF = BD = BC

解:∵∠DAB+∠BAF=∠CAF+∠BAF=90°∠DAB=∠CAF.∵AB=AC,AD=AF.ΔABD≌ΔACF∠ADB=∠AFC.∵∠ABC=∠ADB+∠DAB=45°=∠AFC+∠CAF∠ACF=∠ABD=135°.∵∠ACB=45°.∴∠DCF=90°.∵OD=OFOC=OD=OF.又∵正方形ADEF的边长=2OD=OF=OC=√2.(本题用四点共圆更简单)

1、∵AB=AC ∠ACB=45°∴△ABC是等腰直角三角形且∠BAC=90° 3ACB=45°∵四边形ADEF是正方形∴AD=AF ∠DAF=90°∴∠BAD+∠DAC=∠DAC+∠CAF即∠BAD=∠CAF在△ABD和△ACF中AD=AF AB=AC ∠BAD=∠CAF∴△ABD≌△ACF∴∠A

(接问题)以A为圆心画圆最短半径为(1÷根号2)AB,D在BC中点,

(1)证明:①∵∠BAC=90°,AB=AC, ∴∠ABC=∠ACB=45°, ∵四边形ADEF是正方形, ∴AD=AF,∠DAF=90°, ∵∠BAC=∠BAD+∠DAC=90°, ∠DAF=∠CAF+∠DAC=90°, ∴∠BAD=∠CAF, 在△BAD和△CAF中, AB=AC ∠BAD=∠CAF

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