Chứng minh
.Cho tam giác ABC nhọn có góc A=60*;các đường cao BD,CE.
CMR diện tích ADE=SABC/4
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NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 6 2022
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15 tháng 6 2022
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
30 tháng 9 2021
Tam giác ABD vuông tại D có \(\cos\widehat{A}=\cos60^0=\dfrac{AD}{AB}=\dfrac{1}{2}\)
Tam giác AEC vuông tại E có \(\cos\widehat{A}=\cos60^0=\dfrac{AE}{AC}=\dfrac{1}{2}\)
Ta có \(\left\{{}\begin{matrix}\dfrac{AD}{AB}=\dfrac{AE}{AC}\left(=\dfrac{1}{2}\right)\\\widehat{A}.chung\end{matrix}\right.\Rightarrow\Delta ADE\sim\Delta ABC\left(c.g.c\right)\)
\(\Rightarrow\dfrac{DE}{BC}=\dfrac{AD}{AB}=\dfrac{1}{2}\\ \Rightarrow2DE=BC\)
30 tháng 9 2021
Bạn tự vẽ hình
Đặt \(AB=x\)
Xét \(\Delta DAB\) vuông tại D, ta có:
\(\cos A=\dfrac{AD}{AB}\) (tỉ số lượng giác)
\(\Rightarrow AD=AB.\cos A=x.\cos60^o=0,5x\)
Xét \(\Delta ADB\) và \(\Delta AEC\), ta có:
\(\left\{{}\begin{matrix}\widehat{A}chung\\\widehat{ABD}=\widehat{ACE\left(2gocphunhau\right)}\end{matrix}\right.\)
\(\Rightarrow\Delta ADB\sim\Delta AEC\left(g.g\right)\)
Xét \(\Delta ABC\) và \(\Delta ADE\), ta có:
\(\left\{{}\begin{matrix}\widehat{A}chung\\\dfrac{AB}{AC}=\dfrac{AD}{AE}\left(\Delta ABD\sim\Delta ADE\right)\end{matrix}\right.\)
\(\Rightarrow\Delta ABC\sim\Delta ADE\left(c.g.c\right)\\ \Rightarrow\dfrac{AB}{AD}=\dfrac{BC}{DE}\\ \Rightarrow\dfrac{x}{0,5x}=\dfrac{BC}{DE}\\ \Rightarrow BC=\dfrac{DE.x}{0,5x}=2DE\)
NL
Nguyễn Lê Phước Thịnh
CTVHS
15 tháng 4
a: Gọi IK là phân giác của góc BIC(K∈BC)
Xét ΔABC có \(\hat{BAC}+\hat{ABC}+\hat{ACB}=180^0\)
=>\(\hat{ABC}+\hat{ACB}=180^0-60^0=120^0\)
=>\(2\left(\hat{IBC}+\hat{ICB}\right)=120^0\)
=>\(\hat{IBC}+\hat{ICB}=\frac{120^0}{2}=60^0\)
Xét ΔIBC có \(\hat{IBC}+\hat{ICB}+\hat{BIC}=180^0\)
=>\(\hat{BIC}=180^0-60^0=120^0\)
Ta có: \(\hat{BIE}+\hat{BIC}=180^0\) (hai góc kề bù)
\(\hat{CID}+\hat{BIC}=180^0\) (hai góc kề bù)
Do đó: \(\hat{BIE}=\hat{CID}=180^0-120^0=60^0\)
IK là phân giác của góc BIC
=>\(\hat{BIK}=\hat{CIK}=\frac12\cdot120^0=60^0\)
Xet ΔBEI và ΔBKI có
\(\hat{EBI}=\hat{KBI}\)
BI chung
\(\hat{EIB}=\hat{KIB}\left(=60^0\right)\)
Do đó: ΔBEI=ΔBKI
=>BE=BK và IE=IK
Xét ΔCKI và ΔCDI có
\(\hat{KCI}=\hat{DCI}\)
CI chung
\(\hat{KIC}=\hat{DIC}\)
Do đó: ΔCKI=ΔCDI
=>CK=CD và IK=ID
IE=IK
IK=ID
Do đó: IE=ID
b: BE+CD
=BK+CK
=BC