Bài 2. Cho tam giác nhọn ABC hai đường cao AD và BE cắt nhau tại H. Biết\(\dfrac{HD}{HA}\)=\(\dfrac{1}{2}\). Chứng minh rằng tanB.cotC = 3.
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
A
2 tháng 7 2019
Gọi ( O;R ) , ( I ;r ) lần lượt là các đường tròn ngoại tiếp tam giác ABC, DEF
Tam giác ABC ~ Tam giác DEF ( vì \(\widehat{ABC}=\widehat{DEF};\widehat{BAC}=\widehat{EDF}\)) \(\Rightarrow\widehat{ABC}=\widehat{DEF}\)
\(\widehat{ACB},\widehat{DEF}\)nhọn nên \(\widehat{ACB}=\frac{1}{2}\widehat{AOB};\widehat{DEF}=\frac{1}{2}\widehat{DIE}\)( hệ quả góc nội tiếp )
\(\Rightarrow\widehat{AOB}=\widehat{DIE}\)
\(OA=OB\left(=R\right)\Rightarrow\Delta OAB\)cân tại O
\(ID=IE\left(=r\right)\Rightarrow\Delta IDE\)cân tại I
Do đó Tam giác OAB ~ Tam giác IDE \(\Rightarrow\frac{OA}{ID}=\frac{AB}{DE}\Rightarrow\frac{R}{r}=\frac{3DE}{DE}\)
\(\Rightarrow R=3r\) ( đpcm)
5 tháng 7 2019
Gọi ( O; R ), ( I; R ) lần lượt là các đường tròn ngoại tiếp tam giác ABC, DEF
Tam giác ABC ~ Tam giác DEF ( vì \(\widehat{ABC}=\widehat{DEF;}\widehat{BAC}=\widehat{EDF}\) ) \(\Rightarrow\widehat{ABC}=\widehat{DEF}\)
\(\widehat{ABC}=\widehat{DEF}\)nhọn nên \(\widehat{ACB}=\frac{1}{2}\widehat{AOB};\widehat{DEF}=\frac{1}{2}\widehat{DIE}\)(hệ quả góc nội tiếp )
\(\Rightarrow\widehat{AOB}=\widehat{DIE}\)
\(OA=OA\left(=R\right)\Rightarrow\Delta OAB\)cân tại O
Do đó Tam giác OAB ~ Tam giác IDE\(\Rightarrow\frac{OA}{ID}=\frac{AB}{DE}\Rightarrow\frac{R}{r}=\frac{3DE}{DE}\)
\(\Rightarrow R=3r\left(đpcm\right)\)
Rất vui vì giúp đc bạn <3
NL
Nguyễn Lê Phước Thịnh
CTVHS
17 tháng 5 2021
a) Xét ΔAEB vuông tại E và ΔAFC vuông tại F có
\(\widehat{FAC}\) chung
Do đó: ΔAEB\(\sim\)ΔAFC(g-g)
Suy ra: \(\dfrac{AE}{AF}=\dfrac{AB}{AC}\)(Các cặp cạnh tương ứng tỉ lệ)
hay \(AE\cdot AC=AF\cdot AB\)(ĐPCM)
NL
Nguyễn Lê Phước Thịnh
CTVHS
17 tháng 5 2021
b)
Ta có: \(\dfrac{AE}{AF}=\dfrac{AB}{AC}\)(cmt)
nên \(\dfrac{AE}{AB}=\dfrac{AF}{AC}\)
Xét ΔAEF và ΔABC có
\(\dfrac{AE}{AB}=\dfrac{AF}{AC}\)(cmt)
\(\widehat{FAE}\) chung
Do đó: ΔAEF\(\sim\)ΔABC(c-g-c)
NL
Nguyễn Lê Phước Thịnh
CTVHS
18 tháng 3 2021
a) Xét ΔABE vuông tại E và ΔACF vuông tại F có
\(\widehat{FAC}\) chung
Do đó: ΔABE∼ΔACF(g-g)
b) Ta có: ΔBEC vuông tại E(gt)
nên \(\widehat{EBC}+\widehat{ECB}=90^0\)(hai góc nhọn phụ nhau)
hay \(\widehat{DBH}+\widehat{ACB}=90^0\)(1)
Ta có: ΔDAC vuông tại D(gt)
nên \(\widehat{DAC}+\widehat{DCA}=90^0\)(hai góc nhọn phụ nhau)
hay \(\widehat{DAC}+\widehat{ACB}=90^0\)(2)
Từ (1) và (2) suy ra \(\widehat{DBH}=\widehat{DAC}\)
Xét ΔDBH vuông tại D và ΔDAC vuông tại D có
\(\widehat{DBH}=\widehat{DAC}\)(cmt)
nên ΔDBH\(\sim\)ΔDAC(g-g)
Suy ra: \(\dfrac{DB}{DA}=\dfrac{DH}{DC}\)(Các cặp cạnh tương ứng tỉ lệ)
hay \(DB\cdot DC=DH\cdot DA\)(đpcm)
NQ
Nguyễn Quốc Đạt
CTVVIP
31 tháng 3
a) Chứng minh $\triangle AEB \sim \triangle AFC$
Xét $\triangle ABC$ nhọn với các đường cao $BE$ và $CF$ cắt nhau tại $H$.
Ta có $BE \perp AC$, $CF \perp AB$.
Trong hai tam giác $AEB$ và $AFC$:
- Góc $\widehat{A}$ chung.
- Góc $\widehat{ABE} = \widehat{ACF} = 90^\circ$.
Do đó $\triangle AEB \sim \triangle AFC$ theo trường hợp góc-góc.
b) Chứng minh $\triangle DEF \sim \triangle ABC$
Gọi $D$ là giao điểm của $AH$ với $BC$, $E$ là chân đường cao từ $B$, $F$ là chân đường cao từ $C$.
Xét tam giác $DEF$ và tam giác $ABC$:
- Góc tại $D$ trong $\triangle DEF$ bằng góc tại $A$ trong $\triangle ABC$.
- Góc tại $E$ trong $\triangle DEF$ bằng góc tại $B$ trong $\triangle ABC$.
Do đó $\triangle DEF \sim \triangle ABC$ theo trường hợp góc-góc.
c) Chứng minh $FC$ là tia phân giác góc $DFE$
Xét tam giác $DFE$, với $F$ là chân đường cao từ $C$ và $FC$ cắt góc $DFE$.
Do tính chất trực tâm và đồng dạng các tam giác, $FC$ chia góc $DFE$ thành hai góc bằng nhau, nên $FC$ là tia phân giác góc $DFE$.
NL
Nguyễn Lê Phước Thịnh
CTVHS
15 tháng 5
e: Xét tứ giác BEDC có \(\hat{BEC}=\hat{BDC}=90^0\)
nên BEDC là tứ giác nội tiếp
f: Xét ΔADB vuông tại D và ΔAEC vuông tại E có
\(\hat{DAB}\) chung
Do đó: ΔADB~ΔAEC
=>\(\frac{AD}{AE}=\frac{AB}{AC}\)
=>\(AD\cdot AC=AE\cdot AB\)
g: Xét (O) có
\(\hat{xAC}\) là góc tạo bởi tiếp tuyến Ax và dây cung AC
\(\hat{ABC}\) là góc nội tiếp chắn cung AC
Do đó: \(\hat{xAC}=\hat{ABC}\)
mà \(\hat{ABC}=\hat{ADE}\left(=180^0-\hat{EDC}\right)\)
nên \(\hat{xAC}=\hat{ADE}\)
mà hai góc này là hai góc ở vị trí so le trong
nên Ax//DE
Sửa đề: \(\tan B:\cot C=3\)
Ta có: \(\frac{HD}{HA}=\frac12\)
=>HA=2HD
ta có; AD=AH+HD
=>AD=2HD+HD=3HD
=>\(\frac{DH}{DA}=\frac13\)
Xét ΔADB vuông tại D có \(\tan B=\frac{AD}{DB}\)
Xét ΔBHD vuông tại D có \(\cot BHD=\frac{HD}{DB}\)
mà \(\hat{BHD}=\hat{C}\left(=90^0-\hat{EBC}\right)\)
nên \(\cot C=\frac{HD}{DB}\)
\(\tan B:\cot C=\frac{AD}{DB}:\frac{HD}{DB}=\frac{AD}{HD}=3\)