vẽ tam giác ABC có góc ngoài BCz. Giả sử BCz= 1350 và A=450 . tâm giác ABC là tam giác gì
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6 tháng 2 2020
a,ta có gMAB+gBAC=gMAC
gNAC+gCAB=gNAB
mà gMAB=gNAC=90độ
=>gMAC=gNAB
xét tgMAC và tgNAB có: AM=AB (tgMAB cân tại A)
gMAC=gNAB (cmt)
AN=AC (tgNAC cân tại A)
=> tgMAC = tgNAB (c.g.c)
=>MC=BN (hai cạn tương ứng)
b,gọi AB cắt MC tại H ; gọi MC cắt BN tại I
xét tgAMH vuông tại A => gAMH + gAHM = 90 độ
mà gAHM = gIHB (hai góc đối đỉnh);gAMH = gIBH (vì tgMAC = tgNAB)
=> gIHB+gIBH = 90 độ => gHIB = 90 độ
=>MC vuông góc với BN tại I
c, vì tgABC đều cạnh 4 cm => AB=AC=BC=4 cm
=> AM=AN=4cm
Xét tgAMB vuông tại A,áp dung định lý pytago
=>MB=4 căn 2
tương tự NC=4 căn 2
15 tháng 6 2022
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21 tháng 12 2021
a) Thấy ˆMAC=ˆMAB+ˆBAC=90o+ˆBAC=ˆCAN+ˆBAC=ˆBANMAC^=MAB^+BAC^=90o+BAC^=CAN^+BAC^=BAN^
Từ đây ta xét t/g MAC và BAN ta có:
=>MA=BA; AC=AN
=>ˆMAC=ˆBANMAC^=BAN^
=>ΔMAC=ΔBAN(c−g−c)⇒MC=BNΔMAC=ΔBAN(c−g−c)⇒MC=BN
đpcm.
b)
Ta gọi giao điểm của MC và BN là 1 điểm D
Ta có: ˆDBA=ˆDMA(ΔMAC=ΔBAN(c−g−c))DBA^=DMA^(ΔMAC=ΔBAN(c−g−c))
Nên ˆMBD+ˆBMD=ˆMBA+ˆDBA+ˆBMD=ˆMBA+ˆDMA+ˆBMD=ˆMBAMBD^+BMD^=MBA^+DBA^+BMD^=MBA^+DMA^+BMD^=MBA^
+ˆBMA=90o+BMA^=90o
Xét t/g MBD có ˆMBD+ˆBMD=90o⇒ˆBMD=90oMBD^+BMD^=90o⇒BMD^=90o
⇒BN⊥MC⇒BN⊥MC
Bổ sung D giao điểm nhé vào hình nha bn.
c) Ta giả sử như ABC đều cạnh 4cm (theo đề bài) thì sẽ có: AM=AC=AB=NA=4cm
Áp dụng định lý pi-ta-go ta có:
Cho t/g MAB và NAC thì MB=NC=4√2(cm)42(cm)
Khi ABC đều cạnh 4cm thì AMC = NAB là t/g vuông cân có góc ở đỉnh : 90o+60o=150o
=>ˆAMC=ˆACMAMC^=ACM^= (180o-150o):2=15o
Thì ˆMCB=ˆACB−ˆACM=60o−15o=45oMCB^=ACB^−ACM^=60o−15o=45o
Lại có ˆMAN=360o−90o−60o−90o=120oMAN^=360o−90o−60o−90o=120o
Vì t/gMAN cân tại A nên ˆAMNAMN^= (180o-120o) : 2 =30o
=> ˆCNM=30o+15o=45oCNM^=30o+15o=45o
=>ˆCNM=ˆMCBCNM^=MCB^
=> BC//MN ( so le trong)
đpcm.
21 tháng 12 2021
a) xét tg AMC và tg ABN có
MA=BA(gt)
CA=AN(gt)
ˆMAC=ˆBAN(doˆMAB+ˆBAC=ˆNAC+ˆBAC)MAC^=BAN^(doMAB^+BAC^=NAC^+BAC^)
=>(kết luận)...
b)gọi I là giao điểm của MC và BN
gọi giao điểm của BA và MI là F
vì ΔAMC=ΔABNΔAMC=ΔABNnên
ˆFMA=ˆFBIFMA^=FBI^
mà ˆFMA+ˆFMB=45OFMA^+FMB^=45O
=>ˆFBI+ˆIMB=45OFBI^+IMB^=45O
Xét ΔIMBΔIMBcó góc ˆIMB+ˆMBI+ˆBIMIMB^+MBI^+BIM^= 180O
Mà ˆIMB+ˆMBIIMB^+MBI^=900
21 tháng 12 2021
a) Thấy ˆMAC=ˆMAB+ˆBAC=90o+ˆBAC=ˆCAN+ˆBAC=ˆBANMAC^=MAB^+BAC^=90o+BAC^=CAN^+BAC^=BAN^
Từ đây ta xét t/g MAC và BAN ta có:
=>MA=BA; AC=AN
=>ˆMAC=ˆBANMAC^=BAN^
=>ΔMAC=ΔBAN(c−g−c)⇒MC=BNΔMAC=ΔBAN(c−g−c)⇒MC=BN
đpcm.
b)
Ta gọi giao điểm của MC và BN là 1 điểm D
Ta có: ˆDBA=ˆDMA(ΔMAC=ΔBAN(c−g−c))DBA^=DMA^(ΔMAC=ΔBAN(c−g−c))
Nên ˆMBD+ˆBMD=ˆMBA+ˆDBA+ˆBMD=ˆMBA+ˆDMA+ˆBMD=ˆMBAMBD^+BMD^=MBA^+DBA^+BMD^=MBA^+DMA^+BMD^=MBA^
+ˆBMA=90o+BMA^=90o
Xét t/g MBD có ˆMBD+ˆBMD=90o⇒ˆBMD=90oMBD^+BMD^=90o⇒BMD^=90o
⇒BN⊥MC⇒BN⊥MC
Bổ sung D giao điểm nhé vào hình nha bn.
c) Ta giả sử như ABC đều cạnh 4cm (theo đề bài) thì sẽ có: AM=AC=AB=NA=4cm
Áp dụng định lý pi-ta-go ta có:
Cho t/g MAB và NAC thì MB=NC=4√2(cm)42(cm)
Khi ABC đều cạnh 4cm thì AMC = NAB là t/g vuông cân có góc ở đỉnh : 90o+60o=150o
=>ˆAMC=ˆACMAMC^=ACM^= (180o-150o):2=15o
Thì ˆMCB=ˆACB−ˆACM=60o−15o=45oMCB^=ACB^−ACM^=60o−15o=45o
Lại có ˆMAN=360o−90o−60o−90o=120oMAN^=360o−90o−60o−90o=120o
Vì t/gMAN cân tại A nên ˆAMNAMN^= (180o-120o) : 2 =30o
=> ˆCNM=30o+15o=45oCNM^=30o+15o=45o
=>ˆCNM=ˆMCBCNM^=MCB^
=> BC//MN ( so le trong)
đpcm.
CH
Cô Hoàng Huyền
Admin
VIP
7 tháng 2 2018
a) Ta thấy \(\widehat{MAC}=\widehat{MAB}+\widehat{BAC}=90^o+\widehat{BAC}=\widehat{CAN}+\widehat{BAC}=\widehat{BAN}\)
Xét tam giác MAC và BAN có:
MA = BA
AC = AN
\(\widehat{MAC}=\widehat{BAN}\)
\(\Rightarrow\Delta MAC=\Delta BAN\left(c-g-c\right)\Rightarrow MC=BN\)
b) Gọi giao điểm của MC và BN là J.
Ta có: \(\widehat{JBA}=\widehat{JMA}\)(Vì \(\Delta MAC=\Delta BAN\left(c-g-c\right)\) )
Vậy nên \(\widehat{MBJ}+\widehat{BMJ}=\widehat{MBA}+\widehat{JBA}+\widehat{BMJ}=\widehat{MBA}+\widehat{JMA}+\widehat{BMJ}\)
\(=\widehat{MBA}+\widehat{BMA}=90^o\)
Xét tam giác MBJ có \(\widehat{MBJ}+\widehat{BMJ}=90^o\Rightarrow\widehat{BJM}=90^o\Rightarrow BN\perp MC\)
c) Giả sử tam giác ABC đều cạnh 4 cm thì AB = AC = MA = NA = 4cm
Khi đó áp dụng định lý Pi-ta-go cho tam giác vuông cân MAB và NAC thì \(MB=NC=4\sqrt{2}\left(cm\right)\)
d) Khi tam giác ABC đều cạnh 4cm thì AMC và NAB là các tam giác cân có góc ở đỉnh là: 90o + 60o = 150o
Suy ra \(\widehat{AMC}=\widehat{ACM}=\frac{180^o-150^o}{2}=15^o\)
Vậy thì \(\widehat{MCB}=\widehat{ACB}-\widehat{ACM}=60^o-15^o=45^o\)
Ta có \(\widehat{MAN}=360^o-90^o-60^o-90^o=120^o\)
Tam giác MAN cũng cân tại A nên \(\widehat{AMN}=\frac{180^o-120^o}{2}=30^o\)
\(\Rightarrow\widehat{CMN}=30^o+15^o=45^o\)
Suy ra \(\widehat{CMN}=\widehat{MCB}\)
Chúng lại ở vị trí so le trong nên BC // MN.
21 tháng 12 2021
a) Thấy ˆMAC=ˆMAB+ˆBAC=90o+ˆBAC=ˆCAN+ˆBAC=ˆBANMAC^=MAB^+BAC^=90o+BAC^=CAN^+BAC^=BAN^
Từ đây ta xét t/g MAC và BAN ta có:
=>MA=BA; AC=AN
=>ˆMAC=ˆBANMAC^=BAN^
=>ΔMAC=ΔBAN(c−g−c)⇒MC=BNΔMAC=ΔBAN(c−g−c)⇒MC=BN
đpcm.
b)
Ta gọi giao điểm của MC và BN là 1 điểm D
Ta có: ˆDBA=ˆDMA(ΔMAC=ΔBAN(c−g−c))DBA^=DMA^(ΔMAC=ΔBAN(c−g−c))
Nên ˆMBD+ˆBMD=ˆMBA+ˆDBA+ˆBMD=ˆMBA+ˆDMA+ˆBMD=ˆMBAMBD^+BMD^=MBA^+DBA^+BMD^=MBA^+DMA^+BMD^=MBA^
+ˆBMA=90o+BMA^=90o
Xét t/g MBD có ˆMBD+ˆBMD=90o⇒ˆBMD=90oMBD^+BMD^=90o⇒BMD^=90o
⇒BN⊥MC⇒BN⊥MC
Bổ sung D giao điểm nhé vào hình nha bn.
c) Ta giả sử như ABC đều cạnh 4cm (theo đề bài) thì sẽ có: AM=AC=AB=NA=4cm
Áp dụng định lý pi-ta-go ta có:
Cho t/g MAB và NAC thì MB=NC=4√2(cm)42(cm)
Khi ABC đều cạnh 4cm thì AMC = NAB là t/g vuông cân có góc ở đỉnh : 90o+60o=150o
=>ˆAMC=ˆACMAMC^=ACM^= (180o-150o):2=15o
Thì ˆMCB=ˆACB−ˆACM=60o−15o=45oMCB^=ACB^−ACM^=60o−15o=45o
Lại có ˆMAN=360o−90o−60o−90o=120oMAN^=360o−90o−60o−90o=120o
Vì t/gMAN cân tại A nên ˆAMNAMN^= (180o-120o) : 2 =30o
=> ˆCNM=30o+15o=45oCNM^=30o+15o=45o
=>ˆCNM=ˆMCBCNM^=MCB^
=> BC//MN ( so le trong)
đpcm.
BCz^ + BCA^ = 180o (do BCz^ là góc ngoài của tam giác ABC)
=> BCA^ = 180o - BCz^ = 180o - 135o = 45o
=> BAC^ = BCA^ = 45o
=> Tam giác ABC cân tại B