如图,△ABC中,∠ACB=90°,AC>BC,分别以△ABC的边AB、BC、CA为一边向△ABC外作正方形ABDE、BCMN、CAFG

2025-03-14 23:33:53
推荐回答(3个)
回答1:

辅助线都是延长作高,或直接作高

易证S2=S△ABC

角EAH+∠PAH=90

∠CAB+∠PAH=90

∠EAH=∠CAB

△EHA全等△ACB

EH=CB

又FA=AC

故S△ACB=S1(等低同高)

同理S3=SACB

S1+S2+S3=15

第二题:

同第一题理

GHB≌ABS

GH=AS

MB=BC

等底同高得S2=SABC

同理PDC≌ABC

PD=AS

BC=CN

等底同高得S3=SABC

易证△AVC=△ERA

ER=CV

FA=AB

等底同高得S1=SABC

故S1+S2+S3=3SABC=15

PD=PS

楼主我的答案言简意赅,一目了然,快快采纳

有什么不会继续追问

回答2:

用正弦定理
S2=0.5ab
S1=0.5bc*sinA=SABC
S3=0.5ac*sinB=SABC
所以S1=S2=S3
sinFAE=sin(360-90-90-A)=sin(180-A)=sinA
另外一个同理
正弦定理:a:b:c=sinA:sinB:sinC
三角形面积S=0.5ab*sinC

回答3:

辅助线都是延长作高,或直接作高
易证S2=S△ABC
角EAH+∠PAH=90
∠CAB+∠PAH=90
∠EAH=∠CAB
△EHA全等△ACB
EH=CB
又FA=AC
故S△ACB=S1(等低同高)
同理S3=SACB
S1+S2+S3=15

第二题:
同第一题理
GHB≌ABS
GH=AS
MB=BC
等底同高得S2=SABC
同理PDC≌ABC
PD=AS
BC=CN
等底同高得S3=SABC
易证△AVC=△ERA
ER=CV
FA=AB
等底同高得S1=SABC
故S1+S2+S3=3SABC=15
PD=PS
希望能帮到你!

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