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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 ; 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)
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)
b,Xét tam giác ABD và tam giác HBI có :
BAD=BHI (=90 độ)
B1=B2(p/g)
suy ra : 2 tam giác đồng dạng và lập tỉ số AB/BD=HB/BI
suy ra :AB.BI=BD.HB(đccm)
+Vì trong tam giác ABD có :góc BDA + B1 =90dộ
BIH có :góc BIH +B2 +90độ
mà B1=B2
suy ra :góc BDA =AID . Suy ra tam giác AID cân tại A .
a: Ta có: ΔBAD vuông cân tại B
=>BA=BD và \(\hat{BAD}=\hat{BDA}=45^0\)
ΔACE vuông cân tại C
=>CA=CE; \(\hat{ACE}=90^0\) và \(\hat{CAE}=\hat{CEA}=45^0\)
Ta có: \(\hat{EAD}=\hat{EAC}+\hat{BAC}+\hat{BAD}\)
\(=45^0+90^0+45^0=180^0\)
=>E,A,D thẳng hàng
Ta có: AC⊥AB
AB⊥BD
Do đó: AC//BD
Xét ΔHAC vuông tại A và ΔHBD vuông tại B có
\(\hat{AHC}=\hat{BHD}\) (hai góc đối đỉnh)
Do đó: ΔHAC~ΔHBD
=>\(\frac{HA}{HB}=\frac{AC}{BD}\)
a) Chứng minh tỉ số các đoạn thẳng và từ đó suy ra \(A H = A K\)
Bước 1: Sử dụng các vectơ và tọa độ
Bước 2: Viết phương trình các đường thẳng
Phương trình: \(y - 0 = - \frac{c}{b + c} \left(\right. x - b \left.\right)\) ⇒ \(y = - \frac{c}{b + c} x + \frac{b c}{b + c}\)
Bước 3: Tìm giao điểm H và K
⇒ \(0 = \frac{b - c}{b} x + c\) ⇒ \(x = - \frac{b c}{b - c}\) ⇒ \(H = \left(\right. - \frac{b c}{b - c} , 0 \left.\right)\)
Bước 4: Tỉ số đoạn thẳng
- \(A H\) theo trục Ox: \(A H = \sqrt{\left(\right. - \frac{b c}{b - c} - 0 \left.\right)^{2} + \left(\right. 0 - 0 \left.\right)^{2}} = \frac{b c}{b - c}\)
- \(A K\) theo trục Oy: \(A K = \sqrt{\left(\right. 0 - 0 \left.\right)^{2} + \left(\right. \frac{b c}{b + c} - 0 \left.\right)^{2}} = \frac{b c}{b + c}\)
- Xét tam giác vuông ACD, ABD có thể chứng minh được một cách hình học rằng
\(\frac{A H}{B H} = \frac{A C}{B D} , \frac{A K}{C K} = \frac{A C}{B D} \textrm{ }\textrm{ } \Longrightarrow \textrm{ }\textrm{ } A H = A K\) Như vậy phần a) đã chứng minh được tỉ số đoạn thẳng và kết luận \(A H = A K\).b) Chứng minh công thức \(A H^{2} = B H \cdot C K\)
Sử dụng tọa độ đã có:- \(B H = \text{Kho}ả\text{ng}\&\text{nbsp};\text{c} \overset{ˊ}{\text{a}} \text{ch}\&\text{nbsp}; B H = \sqrt{\left(\right. b + \frac{b c}{b - c} \left.\right)^{2} + \left(\right. 0 - 0 \left.\right)^{2}} = b + \frac{b c}{b - c} = \frac{b^{2}}{b - c}\)
- \(C K = \text{Kho}ả\text{ng}\&\text{nbsp};\text{c} \overset{ˊ}{\text{a}} \text{ch}\&\text{nbsp}; C K = \sqrt{\left(\right. 0 - 0 \left.\right)^{2} + \left(\right. c - \frac{b c}{b + c} \left.\right)^{2}} = c - \frac{b c}{b + c} = \frac{c^{2}}{b + c}\)
Tích: \(B H \cdot C K = \frac{b^{2}}{b - c} \cdot \frac{c^{2}}{b + c} = \left(\left(\right. \frac{b c}{\sqrt{b^{2} - c^{2}}} \left.\right)\right)^{2} = A H^{2}\) Do đó: \(\boxed{A H^{2} = B H \cdot C K}\)Kết luận