Kẻ PD và BE vuông góc AC

Định lý phân giác: \(\dfrac{AN}{NC}=\dfrac{AB}{BC}\Rightarrow\dfrac{AN}{AN+NC}=\dfrac{AB}{AB+BC}\Rightarrow\dfrac{AN}{AC}=\dfrac{AB}{AB+BC}=\dfrac{c}{a+c}\)

Tương tự: \(\dfrac{AP}{AB}=\dfrac{b}{a+b}\)

Talet: \(\dfrac{PD}{BE}=\dfrac{AP}{AB}\)

\(\dfrac{S_{APN}}{S_{ABC}}=\dfrac{\dfrac{1}{2}PD.AN}{\dfrac{1}{2}BE.AC}=\dfrac{AP}{AB}.\dfrac{AN}{AC}=\dfrac{bc}{\left(a+b\right)\left(a+c\right)}\)

Tương tự: \(\dfrac{S_{BPM}}{S_{ABC}}=\dfrac{ac}{\left(a+b\right)\left(b+c\right)}\) ; \(\dfrac{S_{CMN}}{S_{ABC}}=\dfrac{ab}{\left(a+c\right)\left(b+c\right)}\)

\(\Rightarrow\dfrac{S_{APN}+S_{BPM}+S_{CMN}}{S_{ABC}}=\dfrac{bc}{\left(a+b\right)\left(a+c\right)}+\dfrac{ac}{\left(a+b\right)\left(b+c\right)}+\dfrac{ab}{\left(a+c\right)\left(b+c\right)}\)

\(\Rightarrow\dfrac{S_{MNP}}{S_{ABC}}=\dfrac{S_{ABC}-\left(S_{APN}+S_{BPM}+S_{CMN}\right)}{S_{ABC}}=1-\left(\dfrac{bc}{\left(a+b\right)\left(a+c\right)}+\dfrac{ac}{\left(a+b\right)\left(b+c\right)}+\dfrac{ab}{\left(a+c\right)\left(b+c\right)}\right)\)

\(=\dfrac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)

2. Do ABC cân tại C \(\Rightarrow AC=BC=a\)

\(\dfrac{BC}{AB}=k\Rightarrow AB=\dfrac{BC}{k}=\dfrac{a}{k}\)

Do đó:

\(\dfrac{S_{MNP}}{S_{ABC}}=\dfrac{2abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=\dfrac{2.a.a.\dfrac{a}{k}}{2a.\left(a+\dfrac{a}{k}\right)\left(a+\dfrac{a}{k}\right)}=\dfrac{k}{\left(k+1\right)^2}\)

vẽ hình giùm mình với

Không biết vẽ .

\(AH=\dfrac{2S_{ABC}}{BC}=2\sqrt{5}\)

\(\Rightarrow BH=\sqrt{AB^2-AH^2}=\sqrt{5}\)

\(\Rightarrow BH=\dfrac{1}{3}BC\)

\(\Rightarrow\left[{}\begin{matrix}\overrightarrow{BH}=\dfrac{1}{3}\overrightarrow{BC}\\\overrightarrow{BH}=-\dfrac{1}{3}\overrightarrow{BC}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}H\left(1;1\right)\\H\left(3;-3\right)\end{matrix}\right.\) (sử dụng công thức điểm chia đoạn thẳng theo tỉ lệ)

em cảm ơn ạ

198 

cho tớ **** đi

a: Xét ΔABC vuông tại A có \(\tan B=\frac{AC}{AB}=3\)

nên \(\hat{ABC}\) ≃72 độ

ΔABC vuông tại A

=>\(\hat{ABC}+\hat{ACB}=90^0\)

=>\(\hat{ACB}=90^0-72^0=18^0\)

b: Xét ΔABC vuông tại A có AH là đường cao

nên \(BA^2=BH\cdot BC;CA^2=CH\cdot CB\)

=>\(\frac{BA^2}{CA^2}=\frac{BH\cdot CB}{CH\cdot CB}\)

=>\(\frac{HB}{HC}=\left(\frac{AB}{AC}\right)^2=\frac19\)

c: \(\frac{HB}{HC}=\frac19\)

=>\(\frac{HB}{BC}=\frac{1}{10}\)

=>\(S_{ABH}=\frac{1}{10}\cdot S_{ABC}=1,5\left(\operatorname{cm}^2\right)\)