如图,在△ABC中,∠C=90°,点M在BC上,且BM=AC,点N在AC上,且AN=MC.AM与BN相交于点P.求证:∠BPM=45.
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如图,在△ABC中,∠C=90°,点M在BC上,且BM=AC,点N在AC上,且AN=MC.AM与BN相交于点P.求证:∠BPM=45.
如图,在△ABC中,∠C=90°,点M在BC上,且BM=AC,点N在AC上,且AN=MC.AM与BN相交于点P.求证:∠BPM=45.
如图,在△ABC中,∠C=90°,点M在BC上,且BM=AC,点N在AC上,且AN=MC.AM与BN相交于点P.求证:∠BPM=45.
证法一(初中知识证法):
证:已知在△ABC中,∠C=90°,点M在BC上,且BM=AC,点N在AC上,且AN=MC,AM与BN相交于点P.
设AC=BM=X,MC=AN=Y,则
BC=BM+MC=X+Y,CN=AC-AN=X-Y
AM=√(AC^2+MC^2)=√(X^2+Y^2)
过N点作NE⊥AM,交AM于E点,则△AEN∽△ACB
AE/AN=AC/AM,NE/AN=MC/AM
AE=AN*AC/AM=Y*X/√(X^2+Y^2)
NE=AN*MC/AM=Y^2/√(X^2+Y^2)
过P点作PF⊥BC,交BC于F点,则△PFM∽△ACM,△BPF∽△BNC
PF/FM=AC/MC,PF=FM*AC/MC=FM*X/Y
PF/BF=CN/BC,PF=BF*CN/BC=BF*(X-Y)/(X+Y)
BF*(X-Y)/(X+Y)=FM*X/Y
BF=(FM*X/Y)*[(X+Y)/(X-Y)]=FM*X*(X+Y)/[Y*(X-Y)]
BF=BM+FM=X+FM
FM*X*(X+Y)/[Y*(X-Y)]=X+FM
FM=XY*(X-Y)/(X^2+Y^2)
PM/FM=AM/CM
PM=FM*AM/MC=[XY*(X-Y)/(X^2+Y^2)]*[√(X^2+Y^2)/Y]
=X*(X-Y)/√(X^2+Y^2)
PE=AM-AE-PM
=√(X^2+Y^2)-Y*X/√(X^2+Y^2)-X*(X-Y)/√(X^2+Y^2)
=Y^2/√(X^2+Y^2)
=NE
因为NE⊥AM,即NE⊥PE
可知在直角△NEP中,NE=PE
故 ∠EPN=45°
但∠BPM=∠EPN
所以∠BPM=45°
证法二:
证:已知在△ABC中,∠C=90°,点M在BC上,且BM=AC,点N在AC上,且AN=MC,AM与BN相交于点P.
设AC=BM=X,MC=AN=Y,则
BC=BM+MC=X+Y,CN=AC-AN=X-Y
tan∠AMC=AC/MC=X/Y
tan∠NBC=CN/BC=(X-Y)/(X+Y)
∠AMC=∠BPM+∠NBC
∠BPM=∠AMC-∠NBC
tan∠BPM=tan(∠AMC-∠NBC)
=(tan∠AMC-tan∠NBC)/(1+tan∠AMC*tan∠NBC)
=[X/Y-(X-Y)/(X+Y)]/[1+(X/Y)*(X-Y)/(X+Y)]
=[X*(X+Y)-Y*(X-Y)]/[Y*(X+Y)+X*(X-Y)]
=(X ^2+Y ^2)/(X ^2+Y ^2)
=1
因为∠BPM<180°
所以∠BPM=45°
美皮王
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