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Bài 3:
a: Xét ΔHBA vuông tại H và ΔABC vuông tại A có
góc HBA chung
DO đó: ΔHBA\(\sim\)ΔABC
SUy ra: BA/BC=BH/BA
hay \(BA^2=BH\cdot BC\)
b: \(BC=\sqrt{12^2+16^2}=20\left(cm\right)\)
Xét ΔABC có AD là phân giác
nên BD/AB=CD/AC
=>BD/3=CD/4
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{BD}{3}=\dfrac{CD}{4}=\dfrac{BD+CD}{3+4}=\dfrac{20}{7}\)
Do đó: BD=60/7(cm); CD=80/7(cm)
a) Xét \(\Delta ABC\)và \(\Delta HBA\)có:
\(\widehat{B}\) chung
\(\widehat{BAC}=\widehat{BHA}=90^0\)
suy ra: \(\Delta ABC~\Delta HBA\) (g.g)
b) Xét \(\Delta AIH\)và \(\Delta AHB\)có:
\(\widehat{AIH}=\widehat{AHB}=90^0\)
\(\widehat{IAH}\) chung
suy ra: \(\Delta AIH~\Delta AHB\) (g.g)
\(\Rightarrow\)\(\frac{AI}{AH}=\frac{AH}{AB}\) \(\Rightarrow\) \(AI.AB=AH^2\) (1)
Xét \(\Delta AHK\)và \(\Delta ACH\)có:
\(\widehat{HAK}\)chung
\(\widehat{AKH}=\widehat{AHC}=90^0\)
suy ra: \(\Delta AHK~\Delta ACH\) (g.g)
\(\Rightarrow\)\(\frac{AH}{AC}=\frac{AK}{AH}\)
\(\Rightarrow\)\(AK.AC=AH^2\) (2)
Từ (1) và (2) suy ra: \(AI.AB=AK.AC\)
c) \(S_{ABC}=\frac{1}{2}.AH.BC=20\)cm2
Tứ giác \(HIAK\)có: \(\widehat{HIA}=\widehat{IAK}=\widehat{AKH}=90^0\)
\(\Rightarrow\)\(HIAK\)là hình chữ nhật
\(\Rightarrow\)\(AH=IK=4\)cm
Ta có: \(AI.AB=AK.AC\) (câu b)
\(\Rightarrow\)\(\frac{AI}{AC}=\frac{AK}{AB}\)
Xét \(\Delta AIK\)và \(\Delta ACB\)có:
\(\widehat{IAK}\)chung
\(\frac{AI}{AC}=\frac{AK}{AB}\) (cmt)
suy ra: \(\Delta AIK~\Delta ACB\) (c.g.c)
\(\Rightarrow\)\(\frac{S_{AIK}}{S_{ACB}}=\left(\frac{IK}{BC}\right)^2=\frac{4}{25}\)
\(\Rightarrow\)\(S_{AIK}=\frac{4}{25}.S_{ACB}=3,2\)cm2
A ; Ta có : góc ADB=góc AEC=90 độ( đề cho)
góc BAC ( chung)
vậy tam giác ABD đồng dạnh với tam giác ACE ( góc - góc)
B; Xét tam giác EHB và tam giác BCH có:
góc CBH = góc BEH=90 độ
Theo phần a ta lại có góc : EBH=ACE( định lí ta/lét)
vậy suy ra tam giác EHB đồng dạng với tam giác DHC ( góc - góc)
dựa theo 2 tam giác đồng dạng ta có tỉ lệ:
EH/HD=BH/HC ( Ta -lét)
EH*HC=BH*HD( ĐPCM)
C; Theo phần a ta có :
tam giác ABD đồng dạng với tam giác ACE:
suy ra : AB/AD=EA/AC( theo định lí tam giác đồng dạng )
góc A chung
vậy tam giác AED đồng dạng với tam giác ABC ( cạnh -góc -cạnh)
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ABCHKIEF
a)
Xét \(\Delta\)ABC và \(\Delta\)HBA có:
^BAC = ^BHA ( = 90 độ )
^ABC = ^HBA ( ^B chung )
=> \(\Delta\)ABC ~ \(\Delta\)HBA
b) AB = 3cm ; AC = 4cm
Theo định lí pitago ta tính được BC = 5 cm
Từ (a) => \(\frac{AB}{BH}=\frac{BC}{AB}\Rightarrow BH=\frac{AB^2}{BC}=1,8\)m
c) Xét \(\Delta\)AHC và \(\Delta\)AKH có: ^AKH = ^AHC = 90 độ
và ^HAC = ^HAK ( ^A chung )
=> \(\Delta\)AHC ~ \(\Delta\)AKH
=> \(\frac{AH}{AK}=\frac{AC}{AH}\Rightarrow AH^2=AC.AK\)
d) Bạn kiểm tra lại đề nhé!
A B C H 1 2
a) Xét tam giác ABC và tam giác HBA có:
\(\hept{\begin{cases}\widehat{B}chung\\\widehat{BAC}=\widehat{BHA}=90^0\end{cases}\Rightarrow\Delta ABC~\Delta HBA\left(g.g\right)}\)(3)
b) Vì tam giác BHA vuông tại H(gt) nên \(\widehat{B}+\widehat{A1}=90^0\)( 2 góc bù nhau ) (1)
Ta có: \(\widehat{A1}+\widehat{A2}=\widehat{BAC}=90^0\)(2)
(1),(2)\(\Rightarrow\widehat{B}=\widehat{A2}\)
Xét tam giác HBA và tam giác HAC có:
\(\hept{\begin{cases}\widehat{B}=\widehat{A2}\\\widehat{BHA}=\widehat{AHC}=90^0\end{cases}\Rightarrow\Delta HBA~\Delta HAC\left(g.g\right)}\)(4)
\(\Rightarrow\frac{AH}{BH}=\frac{CH}{AH}\)( các đoạn tương ứng tỉ lệ )
\(\Rightarrow AH^2=BH.CH\)(5)
c) Áp dụng định lý Py-ta-go vào tam giác ABC vuông tại A ta có:
\(AB^2+AC^2=BC^2\)
\(\Rightarrow BC=\sqrt{AB^2+AC^2}=10\)(cm)
Từ (3) \(\Rightarrow\frac{AC}{BC}=\frac{AH}{AB}\)( các đoạn tương ứng tỉ lệ )
\(\Rightarrow\frac{8}{10}=\frac{AH}{6}\)
\(\Rightarrow AH=4,8\)(cm)
Từ (4) \(\Rightarrow\frac{HB}{AB}=\frac{HA}{AC}\)
\(\Rightarrow\frac{HB}{6}=\frac{4,8}{8}\)
\(\Rightarrow HB=3,6\)(cm)
Từ (5) \(\Rightarrow HC=6,4\left(cm\right)\)
a) Xét \(\Delta EDC\)và \(\Delta BAC\)
có \(\widehat{EDC}=\widehat{BAC}\left(=90^0\right)\)
\(\widehat{ACB}\)chung
nên \(\Delta EDC\)
\(\Delta BAC\)(g - g)
\(\Rightarrow\frac{EC}{BC}=\frac{CD}{AC}\Rightarrow\frac{EC}{CD}=\frac{BC}{AC}\)
Xét \(\Delta BEC\)và \(\Delta ADC\)
có \(\frac{EC}{CD}=\frac{BC}{AC}\)
\(\widehat{ACB}\)chung
nên \(\Delta BEC\)\(\Delta ADC\)(c - g - c)
Xét \(\Delta AHD\)
ta có AH = HD suy ra \(\Delta AHD\)cân tại H
mà \(\widehat{HAD}=90^0\)nên \(\Delta AHD\)vuông cân tại H
suy ra \(\widehat{ADH}=45^0\)
Gọi giao điểm của AD và BE là O
Xét \(\Delta AOE,\Delta BOD\)
có \(\widehat{OAE}=\widehat{OBD}\)(\(\Delta BEC\)\(\Delta ADC\))
\(\widehat{AOE}=\widehat{BOD}\)(đối đỉnh)
nên \(\Delta AOE\)\(\Delta BOD\)(g - g)
\(\Rightarrow\widehat{AEB}=\widehat{ADH}=45^0\)
Xét \(\Delta ABE\)vuông tại A
có \(\widehat{AEB}=45^0\)nên \(\Delta ABE\)vuông cân tại A
suy ra BE = 2\(\sqrt{AB}\)=\(2\sqrt{2}\)(cm)
b) Gọi giao điểm của AH và BE là I
dễ chứng minh \(\Delta HBA\)\(\Delta ABC\)(g - g)
\(\Rightarrow\frac{AB}{BC}=\frac{BH}{AB}\Rightarrow AB^2=BH\cdot BC\)
có AB = 2 cm, BE = \(2\sqrt{2}\left(cm\right)\)
\(\Rightarrow\frac{AB}{BE}=\frac{1}{\sqrt{2}}\Rightarrow\frac{AB^2}{BE^2}=\frac{1}{2}\Rightarrow\frac{BH\cdot BC}{BE^2}=\frac{1}{2}\)
\(\Rightarrow\frac{BH}{BE}\cdot\frac{BC}{BE}=\frac{1}{2}\Rightarrow\frac{BH}{BE}=\frac{1}{2}\cdot\frac{BE}{BC}\Rightarrow\frac{BH}{BE}=\frac{BM}{BC}\)
Xét \(\Delta BHM\)và \(\Delta BEC\)
có \(\frac{BH}{BE}=\frac{BM}{BC}\)
\(\widehat{EBC}\)chung
nên \(\Delta BHM\)\(\Delta BEC\)(c - g - c)
\(\Rightarrow\widehat{IMH}\left(\widehat{BMH}\right)=\widehat{BCE}\)
mà \(\widehat{BCE}=\widehat{IAB}\)(cùng phụ với góc \(\widehat{B}\))
\(\Rightarrow\widehat{IMH}=\widehat{IAB}\)
dễ cm \(\Delta IAB\)\(\Delta IMH\)(g - g)
\(\Rightarrow\widehat{AHM}\left(\widehat{IHM}\right)=\widehat{IBA}=45^0\)
c) có AK là phân giác \(\Delta ABC\)
nên \(\frac{BK}{KC}=\frac{AB}{AC}\Rightarrow\frac{BK}{KC+BK}=\frac{AB}{AB+AC}\Rightarrow\frac{BK}{BC}=\frac{AB}{AB+AC}\)(1)
dễ cm \(\Delta ABH\)\(\Delta CAH\)(g - g)
\(\Rightarrow\frac{AB}{AC}=\frac{AH}{HC}\Rightarrow\frac{AB}{AB+AC}=\frac{AH}{AH+HC}\Rightarrow\frac{AB}{AB+AC}=\frac{HD}{AH+HC}\)(2)
từ (1) và (2) suy ra
\(\frac{BK}{BC}=\frac{HD}{AH+HC}\)
A B C H I K
a, bạn tự làm nhé
b, Xét tam giác ABH và tam giác CAH ta có
^AHB = ^CHA = 900
^ABH = ^CAH ( cùng phụ ^BAH )
Vậy tam giác ABH ~ tam giác CAH ( g.g )
\(\Rightarrow\frac{AH}{CH}=\frac{BH}{AH}\Rightarrow AH^2=BH.CH\)
c, mình làm hơi tắt nhé, bạn dùng tỉ lệ thức xác định tam giác đồng dạng nhé
Dễ có : \(AH^2=AK.AC\)(1)
\(AH^2=AI.AB\)(2)
Từ (1) ; (2) suy ra : \(AK.AC=AI.AB\Rightarrow\frac{AK}{AB}=\frac{AI}{AC}\)
Xét tam giác AIK và tam giác ACB
^A _ chung
\(\frac{AK}{AB}=\frac{AI}{AC}\)( cmt )
Vậy tam giác AIK ~ tam giác ACB ( c.g.c )
Bạn tự vẽ hình nhaa =)) <3
a) Xét \(\Delta HBA\)và \(\Delta ABC\)có
\(\widehat{ABC}chung\)
\(\widehat{BHA}=\widehat{BAC}\)( vì cùng = 90 độ)
\(\Rightarrow\Delta HBA\)đồng dạng với \(\Delta ABC\)(g.g)
b) Vì \(\Delta ABC\)vuông tại A (gt)
\(\Rightarrow AB^2+AC^2=BC^2\)( định lý Py-ta-go)
thay số vào tính được AB= 20 (cm) nhé
Vì \(\Delta HBA\)đồng đạng với \(\Delta ABC\)(cmt)
\(\Rightarrow\frac{AH}{AC}=\frac{AB}{BC}\)( định nghĩ tam giác đd)
thay số vào rồi tính được AH= 12(cm) nè
c) Xét \(\Delta HCO\)và \(\Delta ACI\)có
\(\widehat{HCO}=\widehat{ACI}\)( vì CI là tia phân giác )
\(\widehat{OHC}=\widehat{IAC}\)( cùng = 90 độ)
\(\Rightarrow\Delta HCO\)đòng dạng với \(\Delta ACI\)(g.g)
\(\Rightarrow\frac{HC}{AC}=\frac{HO}{AI}\)( đn tam giác đd)
\(\Rightarrow HC.AI=AC.HO\)
d) Mình chưa ngĩ ra nhwung mình nghĩ sẽ dựa vào Sabc và tỉ số đồng dạng đó ạ :(((