如图,在三角形ABC中,AB=BC,在BC上取点M,在MC上取点N,使MN=NA,若角BAM=角NAC,则角MAC=

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如图,在三角形ABC中,AB=BC,在BC上取点M,在MC上取点N,使MN=NA,若角BAM=角NAC,则角MAC=如图,在三角形ABC中,AB=BC,在BC上取点M,在MC上取点N,使MN=NA,若

如图,在三角形ABC中,AB=BC,在BC上取点M,在MC上取点N,使MN=NA,若角BAM=角NAC,则角MAC=
如图,在三角形ABC中,AB=BC,在BC上取点M,在MC上取点N,使MN=NA,若角BAM=角NAC,则角MAC=

如图,在三角形ABC中,AB=BC,在BC上取点M,在MC上取点N,使MN=NA,若角BAM=角NAC,则角MAC=
角MAC=角MAN+角NAC=角AMN+角NAC=角B+角BAM+角NAC=2*角NAC+角B=2*角NAC+(180°-2*角C)=2*角NAC+(180°-2(角BAM+MAC))=2*角NAC+180°-2*角NAC-2*角MAC=180°-2*角MAC
整理两边,3*角MAC=180°,因此角MAC=60°
解毕

设角BAM=角CAN=θ,角NAM=角NMA=α,则角MAC=α+θ
且角ANM=180°-2α,故角C=180°-2α-θ,又角C=角BAC,即180°-2α-θ=α+2θ
所以α+θ=60°。即为所求。