[紧急]一道初一数学几何题如图,∠CAD=3∠BAD ,∠ABE=3∠CBE ,∠BCF=3∠ACF ,BE、CF交于点M ,CF、AD交于点N ,且∠BMF=2∠CND ,求∠BAC .这个图片更清楚:
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![[紧急]一道初一数学几何题如图,∠CAD=3∠BAD ,∠ABE=3∠CBE ,∠BCF=3∠ACF ,BE、CF交于点M ,CF、AD交于点N ,且∠BMF=2∠CND ,求∠BAC .这个图片更清楚:](/uploads/image/z/3642712-16-2.jpg?t=%5B%E7%B4%A7%E6%80%A5%5D%E4%B8%80%E9%81%93%E5%88%9D%E4%B8%80%E6%95%B0%E5%AD%A6%E5%87%A0%E4%BD%95%E9%A2%98%E5%A6%82%E5%9B%BE%2C%E2%88%A0CAD%3D3%E2%88%A0BAD+%2C%E2%88%A0ABE%3D3%E2%88%A0CBE+%2C%E2%88%A0BCF%3D3%E2%88%A0ACF+%2CBE%E3%80%81CF%E4%BA%A4%E4%BA%8E%E7%82%B9M+%2CCF%E3%80%81AD%E4%BA%A4%E4%BA%8E%E7%82%B9N+%2C%E4%B8%94%E2%88%A0BMF%3D2%E2%88%A0CND+%2C%E6%B1%82%E2%88%A0BAC+.%E8%BF%99%E4%B8%AA%E5%9B%BE%E7%89%87%E6%9B%B4%E6%B8%85%E6%A5%9A%EF%BC%9A)
[紧急]一道初一数学几何题如图,∠CAD=3∠BAD ,∠ABE=3∠CBE ,∠BCF=3∠ACF ,BE、CF交于点M ,CF、AD交于点N ,且∠BMF=2∠CND ,求∠BAC .这个图片更清楚:
[紧急]一道初一数学几何题
如图,∠CAD=3∠BAD ,∠ABE=3∠CBE ,∠BCF=3∠ACF ,
BE、CF交于点M ,CF、AD交于点N ,
且∠BMF=2∠CND ,求∠BAC .
这个图片更清楚:
[紧急]一道初一数学几何题如图,∠CAD=3∠BAD ,∠ABE=3∠CBE ,∠BCF=3∠ACF ,BE、CF交于点M ,CF、AD交于点N ,且∠BMF=2∠CND ,求∠BAC .这个图片更清楚:
∠BMF= ∠BCF+∠CBE = 2∠CND = 2(∠CAD+∠ACF)
即得出∠BCF+∠CBE = 2(∠CAD+∠ACF)
而∠BCF = 3∠ACB /4 (即4分之3大小的∠ACB)
∠CBE = ∠ABC /4
∠CAD = 3∠BAC /4
∠ACF = ∠ACB /4
由上可得出3∠ACB /4 + ∠ABC /4 = 2(3∠BAC /4 + ∠ACB /4)
由于∠BAC + ∠ABC + ∠ACB = 180
根据上面两个式子可以计算出∠BAC=180/7(7分之180度).
∵∠BMF=1/4∠ABC+3/4∠BCA,∠CND=3/4∠BAC+1/4∠BCA,∠BMF=2∠CND
∴1/4∠ABC+3/4∠BCA=2(3/4∠BAC+1/4∠BCA)
∴∠ABC+3∠BCA=6∠BAC+2∠BCA
∴∠ABC+∠BCA=6∠BAC
又∵∠ABC+∠BCA+∠BAC=180°
∴7∠BAC=180°∴∠BAC=180°/7
简单啊!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
∵∠BMF=1/4∠ABC+3/4∠BCA,∠CND=3/4∠BAC+1/4∠BCA,∠BMF=2∠CND
∴1/4∠ABC+3/4∠BCA=2(3/4∠BAC+1/4∠BCA)
∴∠ABC+3∠BCA=6∠BAC+2∠BCA
∴...
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简单啊!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
∵∠BMF=1/4∠ABC+3/4∠BCA,∠CND=3/4∠BAC+1/4∠BCA,∠BMF=2∠CND
∴1/4∠ABC+3/4∠BCA=2(3/4∠BAC+1/4∠BCA)
∴∠ABC+3∠BCA=6∠BAC+2∠BCA
∴∠ABC+∠BCA=6∠BAC
又∵∠ABC+∠BCA+∠BAC=180°
∴7∠BAC=180°∴∠BAC=180°/7 =25.5度
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