Đáp án:
Giải thích các bước giải:
Bai 1
a) Xet tam giac ABC can tai A
Ta co :AH vuong goc voi BC
-> AH la duong cao cua tam giac ABC
Ma trong tam giac can duong cao dong thoi la duong trung tuyen
->HB=HC
b)Xet tam giac ABC can tai A
Ta co :HB=HC(cmt)
->BC=2HB
->HB=BC/2
->HB=8/2=4
Xet tam giac AHB vuong tai H
Ta co: AB^2 = AH^2 + HB^2 (dinh li Pi ta go)
-> AH^2=AB^2 - HB^2
-> AH^2 =5^2 - 4^2 =9
-> AH =3
c) Xet tam giac ABC can tai A
Ta co: AB=AC (gt)
->BD=CE
Xet tam giac DBH va tam giac ECH
Ta co:BD=CE (cmt)
gocB = gocC (tinh chat cua tam giac can)
HB=HC(cmt)
->tam giac DBH = tam giac ECH (c-g-c)
->DH=EH ( 2 canh tuong ung )
Xet tam giac HDE
Ta co:DH=EH (cmt)
->tam giac HDE la tam giac can tai H
Bai 2
Xet tam giac ABC can tai A
Ta co:
AH la duong cao cua tam giac ABC
Ma trong tam giac can duong cao dong thoi la duong trung tuyen , cung dong thoi la duong phan giac
->BH=HC
->AH la duong phan giac cua tam giac ABC
Xet tam giac AHB va tam giac AHC
Ta co: AH la canh chung
gocAHB = gocAHC =90o
BH=HC (cmt)
->tam giac AHB = tam giac AHC ( c-g-c)
b) Xet tam giac AHB vuong tai H
Ta co:
AB^2 = AH^2 + HB^2 (dinh li Pi ta go)
-> AH^2=AB^2 - HB^2
-> AH^2 =10^2 - 8^2 =36
-> AH=6
Bai 3:
a) Xet tam giac ABC
Ta co: CA=CB=10 (gt)
->tam giac ABC la tam giac can tai C
Ta co: CI vuong goc voi AB
-> CI la duong cao cua tam giac ABC
Ma trong tam giac can duong cao dong thoi la duong trung tuyen
->IB=IA
Xet tam giac ABC can tai C
Ta co :IB=IA( cmt)
->AB=2IB
->IB=AB/2
->IB=12/2=6
Xet tam giac BIC vuong tai I
Ta co :
BC^2 = IC^2 + IB^2 (dinh li Pi ta go)
-> IC^2=BC^2 - IB^2
-> IC^2 =10^2 - 6^2 =64
-> IC=8
b) Xet tam giac ABC can tai C
Ta co :CA=CB (gt)
->HA=KB
Xet tam giac AHI va tam giac BKI
Ta co :gocA=gocB(tinh chat cua tam giac can)
HA=KB (cmt)
gocH=gocK=90o
-> tam giac AHI = tam giac BKI (g-c-g)
-> HI =KI
c)
Xet tam giac ABC can tai C
Ta co :CA=CB (gt)
->HC=KC
Xet tam giac HCK
Ta co :HC=KC (cmt)
->tam giac HCK can tai C
Ma gocH =gocA
->HK//AB
Bai 4
a)
Xet tam giac AHB vuong tai H
Ta co: AB^2 = AH^2 + HB^2 (dinh li Pi ta go)
-> AH^2=AB^2 - HB^2
-> AH^2 =10^2 - 6^2 =64
-> AH =8
b)
Xet tam giac AHB va tam giac AHC
Ta co: AH la canh chung
gocAHB = gocAHC =90o
BA=CA(tinh chat cua tam giac can)
->tam giac AHB = tam giac AHC ( c-g-c)
c)
Xet tam giac ABC can tai A
Ta co :AH vuong goc voi BC
-> AH la duong cao cua tam giac ABC
Ma trong tam giac can duong cao dong thoi la duong trung tuyen
->HB=HC
Ta co: AB=AC (gt)
->BD=CE
Xet tam giac DBH va tam giac ECH
Ta co:BD=CE (cmt)
gocB = gocC (tinh chat cua tam giac can)
HB=HC(cmt)
->tam giac DBH = tam giac ECH (c-g-c)
->DH=EH ( 2 canh tuong ung )
Xet tam giac HDE
Ta co:DH=EH (cmt)
->tam giac HDE la tam giac can tai H
d)
Xet tam giac ABC can tai A
Ta co: AB=AC (gt)
->DA=EA
Xet tam giac DAE
Ta co:DA=EA (cmt)
-> tam giac DAE can tai A
Xet tam giac DAE can tai A
Ta co :AH vuong goc voi DE
-> AH la duong cao cua tam giac DAE
Ma trong tam giac can duong cao dong thoi la duong trung truc
->AH la duong trung truc cua DE
Bai 5
a)Xet tam giac ADB va tam giac ADC
Ta co: BD=DC(D la trung dien cua BC)
gocB=gocC (tinh chat cua tam giac can)
BA=CA(tinh chat cua tam giac can)
->tam giac AHB = tam giac AHC ( c-g-c)
b)
Xet tam giac ABC can tai A
Ta co : D la trung dien cua BC
-> AD la duong trung tuyen cua tam giac ABC
Ma trong tam giac can duong trung tuyen dong thoi la duong cao
->AD la duong cao cua BC
->AD vuong goc voi BC
c)
Xet tam giac ABC can tai A
Ta co :DB=DC(cmt)
->BC=2DC
->DC=BC/2
->DC=12/2=6
Xet tam giac ADC vuong tai D
Ta co: AC^2 = AD^2 + DC^2 (dinh li Pi ta go)
-> AD^2=AC^2 - DC^2
-> AD^2 =10^2 - 6^2 =64
-> AD =8
d)
Xet tam giac ABC can tai A
Ta co: AB=AC (gt)
->BE=CF
Xet tam giac BED va tam giac CFD
Ta co:BD=DC (cmt)
gocB = gocC (tinh chat cua tam giac can)
BE=CF (cmt)
->tam giac BED = tam giac CFD (c-g-c)
->DE=DF ( 2 canh tuong ung )
Xet tam giac EDF
Ta co:DE=DF (cmt)
->tam giac EDF la tam giac can tai D
