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a) Xét tam giác ABC và tam giác HBA có Góc ABC chungg,góc BHA=góc BAC=90 độ
=> Tam giác ABC đồng dạng với tam giác HBA(gg)=> \(\frac{AB}{HB}=\frac{BC}{AB}\)=> AB^2=BH.BC
b)Tam giác ABC có BF là phân giác góc ABC=>\(\frac{BC}{AB}=\frac{FC}{AF}\)mà \(\frac{AB}{HB}=\frac{BC}{AB}\)=>\(\frac{AB}{BH}=\frac{FC}{AF}\left(1\right)\)
Tam giác ABH có BE là phân giác goc ABH =>\(\frac{BA}{BH}=\frac{AE}{EH}\left(2\right)\)
Từ 1 và 2=>\(\frac{FC}{AF}=\frac{AE}{EH}=>\frac{EH}{AE}=\frac{AF}{FC}\)
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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) 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) Xét hai tam giác vuông: \(\Delta ABC\) và \(\Delta HBA\) có:
\(\widehat{B}\) chung
\(\Rightarrow\Delta ABC\) ∽\(\Delta HBA\left(g-g\right)\)
\(\Rightarrow\dfrac{AC}{AH}=\dfrac{BC}{AB}\)
\(\Delta ABC\) vuông tại A (gt)
\(\Rightarrow BC^2=AB^2+AC^2\left(Pythagore\right)\)
\(=9^2+12^2\)
\(=225\)
\(\Rightarrow BC=15\left(cm\right)\)
\(\Rightarrow\dfrac{12}{AH}=\dfrac{15}{9}\)
\(\Rightarrow AH=\dfrac{9.12}{15}=7,2\left(cm\right)\)
b) Xét hai tam giác vuông: \(\Delta AHB\) và \(\Delta CHA\) có:
\(\widehat{BAH}=\widehat{ACH}\) (cùng phụ \(\widehat{ABC}\))
\(\Rightarrow\Delta AHB\) ∽\(\Delta CHA\left(g-g\right)\)
\(\Rightarrow\dfrac{AH}{CH}=\dfrac{HB}{AH}\)
\(\Rightarrow AH^2=HB.HC\)
c) Do \(\Delta ABC\) ∽\(\Delta HBA\left(cmt\right)\)
\(\Rightarrow\dfrac{AB}{HB}=\dfrac{BC}{AB}\)
\(\Rightarrow HB=\dfrac{AB^2}{BC}=\dfrac{9^2}{15}=5,4\left(cm\right)\)
Do \(BE\) là tia phân giác của \(\widehat{ABC}\) (gt)
\(\Rightarrow\widehat{ABE}=\widehat{CBE}\)
\(\Rightarrow\widehat{ABE}=\widehat{HBF}\)
Xét hai tam giác vuông: \(\Delta ABE\) và \(\Delta HBF\) có:
\(\widehat{ABE}=\widehat{HBF}\left(cmt\right)\)
\(\Rightarrow\Delta ABE\) ∽\(\Delta HBF\left(g-g\right)\)
\(\Rightarrow\dfrac{S_{ABE}}{S_{HBF}}=\left(\dfrac{AB}{HB}\right)^2=\left(\dfrac{9}{7,2}\right)^2=\dfrac{25}{16}\)