如图,在三角形ABC中,角C=90°,点M在BC上,且BM=AC,N在AC上,且AN=MC,AM与BN相交于P,求证角B
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/07/14 04:56:28
如图,在三角形ABC中,角C=90°,点M在BC上,且BM=AC,N在AC上,且AN=MC,AM与BN相交于P,求证角BPM=45°
急要,17:00之前就要回复
将会奉献上我所有的分
有更简洁的吗
急要,17:00之前就要回复
将会奉献上我所有的分
有更简洁的吗
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
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
如图,在三角形ABC中,角C=90°,点M在BC上,且BM=AC,N在AC上,且AN=MC,AM与BN相交于P,求证角B
如图,在三角形ABC中,∠C=90°,点M在BC上,且BM=AC,N在AC上,且AN=MC.AM与BN相交于点P.求证:
如图,在三角形ABC中,角C=90度,点M在BC上,且BM=AC,点N在AC上,且AN=MC.AM.BN相交于点P.求证
在△ABC中,∠C=90°,点M在BC上,且BM=AC,N在AC上,且AN=CM,AM于BN相交于点P,求证:∠BMP=
直角三角形ABC:角C=90°,N点在AC上,M点在BC上,且AN=MC,BM=AC,连接BN与AM交P点.试求:角BP
在直角三角形ABC中,角C=90度,点M在BC上,BM=AC,AN=MC,AM与BN相交于点P,求证角BPM=45度
如图b,△ABC为等边三角形,M在直线BC上,N在直线AC上,且BM=CN,直线BN与AM相交于Q点(1)图a中,∠BQ
RtABC三角形中,角ABC=90 m是BC的中点,N在边AC上,且AN=2NC,AM与BN相交于点P ,求AP:PM(
在三角形ABC中,点M在BC的中点,点N在边AC上,且AN=2NC,AM与BN相交于点P,求向量AP比向量PM的值.
在三角形ABC中,点M是BC的中点,点N在AC上,且AN=2NC,AM与BN相交于点P,求AP:PM的值
平面向量已知在三角形ABC中,M是BC的中点,点N在边AC上,且Ac=2NC,AM与BN相交于P,求AP与PM的比值.图
如图,在RT三角形ABC中,点M是斜边BC的中点,点N在边AC上,且AN=2NC,AM与BM交与点P,求AP:PM的值