Cho tam giác nhọn ABC có BC = a ; CA = b; AB = c . CMR :
\(sin\frac{A}{2}\le\frac{a}{2\sqrt{bc}}\)
xác định dạng của ABC khi dấu " = " xảy ra
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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)
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)
a: Xét ΔAKB và ΔAKC có
AB=AC
góc BAK=góc CAK
AK chung
=>ΔAKB=ΔAKC
ΔABC cân tại A
mà AK là phân giác
nên AK vuông góc CB
b: Xét ΔACB có
BM,AK là trung tuyến
BM cắt AK tại G
=>G là trọng tâm
c: BK=CK=18/2=9cm
=>\(AK=\sqrt{30^2-9^2}=3\sqrt{91}\left(cm\right)\)
=>\(AG=2\sqrt{91}\left(cm\right)\)
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- Dựng phân giác AD của góc A . Sau đó dựng BB' và CC' vuông góc với AD
- Đặt BB' = x , CC' = y . Ta có :
+) \(\Delta ABB'\)cân tại A \(sin\frac{A}{2}=\frac{x}{2c}\)
+) \(\Delta ACC'\)cân tại A \(sin\frac{A}{2}=\frac{y}{2b}\)
\(\Rightarrow sin^2\frac{A}{2}=\frac{xy}{4bc}\)
Để cm(1) , ta cần cm : \(xy\le a^2\)
+) Trong tam giác BHD vuông tại H ta có : \(BH\le CD\)hay \(\frac{x}{2}\le BD\)
+) Trong tam giác CKD vuông tại K ta có : \(CK\le CH\)hay \(\frac{y}{2}\le CD\)
\(\Rightarrow a=BD+CD\ge\frac{x+y}{2}\ge\sqrt{xy}\)
\(\Rightarrow a^2\ge xy\left(đpcm\right)\)
Kẻ phân giác AD của tam giác ABC (D nằm trên đoạn BC)
Từ B,C kẻ các đường vuông góc với đường thẳng AD tại E,F
Khi đó ta có: \(\sin\widehat{BAE}=\frac{BE}{AB}=\frac{BE}{c}\) ; \(\sin\widehat{FAC}=\frac{CF}{AC}=\frac{CF}{b}\)
Mà \(\sin\frac{\widehat{A}}{2}=\sin\widehat{BAE}=\sin\widehat{FAC}=\frac{BE}{c}=\frac{CF}{b}=\frac{BE+CF}{b+c}\)
Ta thấy \(\hept{\begin{cases}BE\le BD\\CF\le CD\end{cases}}\Rightarrow BE+CF\le BD+CD=BC\)
Lại có theo bất đẳng thức Cauchy: \(b+c\ge2\sqrt{bc}\)
\(\Rightarrow\sin\frac{\widehat{A}}{2}=\frac{BE+CF}{b+c}\le\frac{BC}{2\sqrt{bc}}=\frac{a}{2\sqrt{bc}}\)
Dấu "=" xảy ra khi tam giác ABC cân tại A