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\(\dfrac{sin\left(a-b\right)}{sina.sinb}+\dfrac{sin\left(b-c\right)}{sinb.sinc}+\dfrac{sin\left(c-a\right)}{sinc.sina}\)
\(=\dfrac{sina.cosb-cosa.sinb}{sina.sinb}+\dfrac{sinb.cosc-cosb.sinc}{sinb.sinc}+\dfrac{sinc.cosa-cosc.sina}{sina.sinc}\)
\(=\dfrac{cosb}{sinb}-\dfrac{cosa}{sina}+\dfrac{cosc}{sincc}-\dfrac{cosb}{sinb}+\dfrac{cosa}{sina}-\dfrac{cosc}{sincc}\)
\(=0\)
a)
\(\overrightarrow{AK}=\overrightarrow{AI}+\overrightarrow{IK}=\overrightarrow{AI}+\dfrac{1}{2}\overrightarrow{IB}=\overrightarrow{AI}+\dfrac{1}{2}\left(\overrightarrow{IA}+\overrightarrow{AB}\right)\)
\(=\overrightarrow{AI}+\dfrac{1}{2}\overrightarrow{IA}+\dfrac{1}{2}\overrightarrow{AB}\)\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}\).
b) Theo câu a:
\(\overrightarrow{AK}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}\overrightarrow{AI}=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{2}.\dfrac{1}{2}\left(\overrightarrow{AB}+\overrightarrow{AC}\right)\)
\(=\dfrac{1}{2}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}=\dfrac{3}{4}\overrightarrow{AB}+\dfrac{1}{4}\overrightarrow{AC}\).
Câu 3:
\(\left|\overrightarrow{AC}+\overrightarrow{AH}\right|=\sqrt{AC^2+AH^2+2\cdot AC\cdot AH\cdot cos30}\)
\(=\sqrt{a^2+\left(\dfrac{a\sqrt{3}}{2}\right)^2+2\cdot a\cdot\dfrac{a\sqrt{3}}{2}\cdot\dfrac{\sqrt{3}}{2}}\)
\(=\sqrt{a^2+\dfrac{3}{4}a^2+\dfrac{3a^2}{4}}=\dfrac{\sqrt{7}}{2}a\)
Thay vì \(\alpha;\beta;\gamma\) khó gõ kí tự, mình chuyển thành \(a,b,c\) cho dễ, bạn tự thay lại.
Do ABCD là hbh \(\Rightarrow\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}\)
- Chứng minh chiều thuận: I, F, K thẳng hàng \(\Rightarrow\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\)
Do I, F, K thẳng hàng \(\Rightarrow\) tồn tại một số \(k\ne0\) để \(\overrightarrow{KF}=k.\overrightarrow{KI}\)
\(\Rightarrow\left(\overrightarrow{KA}+\overrightarrow{AF}\right)=k.\left(\overrightarrow{KA}+\overrightarrow{AI}\right)\Rightarrow\left(-c.\overrightarrow{AD}+b.\overrightarrow{AC}\right)=k\left(-c.\overrightarrow{AD}+a.\overrightarrow{AB}\right)\)
\(\Rightarrow\overrightarrow{AD}\left(ck-c\right)=k.a.\overrightarrow{AB}-b.\overrightarrow{AC}=ka.\overrightarrow{AB}-b.\overrightarrow{AB}-b.\overrightarrow{AD}\)
\(\Rightarrow\overrightarrow{AD}\left(ck-c+b\right)=\overrightarrow{AB}\left(ka-b\right)\) (1)
Do \(\overrightarrow{AD};\overrightarrow{AB}\) không cùng phương \(\Rightarrow\left(1\right)\) xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}ck-c+b=0\\ka-b=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\dfrac{c-b}{c}\\k=\dfrac{b}{a}\end{matrix}\right.\)
\(\Rightarrow\dfrac{c-b}{c}=\dfrac{b}{a}\Rightarrow1=\dfrac{b}{a}+\dfrac{b}{c}\Rightarrow\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\) (đpcm)
- Chứng minh chiều nghịch: \(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\Rightarrow\) I, F, K thẳng hàng
\(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\Rightarrow b=\dfrac{ac}{a+c}\)
\(\overrightarrow{FI}=\overrightarrow{FA}+\overrightarrow{AI}=-b.\overrightarrow{AC}+a.\overrightarrow{AB}=-b\left(\overrightarrow{AB}+\overrightarrow{AD}\right)+a.\overrightarrow{AB}\)
\(\Rightarrow\overrightarrow{FI}=-\dfrac{ac}{a+c}\overrightarrow{AB}-\dfrac{ac}{a+c}\overrightarrow{AD}+a.\overrightarrow{AB}=\dfrac{a^2}{a+c}\overrightarrow{AB}-\dfrac{ac}{a+c}\overrightarrow{AD}\)
\(\Rightarrow\overrightarrow{FI}=\dfrac{a}{a+c}\left(a.\overrightarrow{AB}-c.\overrightarrow{AD}\right)\) (1)
Lại có \(\overrightarrow{KI}=\overrightarrow{KA}+\overrightarrow{AI}=-c.\overrightarrow{AD}+a.\overrightarrow{AB}=a.\overrightarrow{AB}-c.\overrightarrow{AD}\) (2)
Từ (1), (2) \(\Rightarrow\overrightarrow{FI}=\dfrac{a}{a+c}\overrightarrow{KI}\) ; mà \(\dfrac{a}{a+c}\) là hằng số \(\ne0\)
\(\Rightarrow F,I,K\) thẳng hàng (đpcm)
Vậy F, I, K thẳng hàng khi và chỉ khi \(\dfrac{1}{b}=\dfrac{1}{a}+\dfrac{1}{c}\)