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A B C I F G H x y z
dat HI=x, HF=y, HG=z
ta co \(\frac{SBHC}{SABC}=\frac{\frac{1}{2}.HI.BC}{\frac{1}{2}AI.BC}=\frac{HI}{AI}=\) \(\frac{x}{x+8}\)
ttu \(\frac{SAHC}{SABC}=\frac{y}{y+\sqrt{14}}\) \(\frac{SHAB}{SABC}=\frac{z}{z+\sqrt{44}}\)
cộng vế vs vế \(\frac{x}{x+8}+\frac{y}{y+\sqrt{14}}+\frac{z}{z+\sqrt{44}}=\frac{SHBC+SHAC+SHAB}{SABC}=1\) (1)
do \(\Delta AHF\simeq\Delta BHI\rightarrow\frac{HF}{HI}=\frac{y}{x}=\frac{AH}{BH}=\frac{8}{\sqrt{14}}\Rightarrow y=\frac{8}{\sqrt{14}}x\)
ttu \(\Delta AHG\simeq\Delta CHI\Rightarrow z=\frac{8}{\sqrt{44}}x\)
the vao 1 ta co \(\frac{x}{x+8}+\frac{\frac{8}{\sqrt{14}}x}{\frac{8}{\sqrt{14}}x+\sqrt{14}}+\frac{\frac{8x}{\sqrt{44}}}{\frac{8x}{\sqrt{44}}+\sqrt{44}}=1\)
\(\Leftrightarrow\frac{x}{x+8}+\frac{8x}{8x+14}+\frac{8x}{8x+44}=1\)
giải ra bn có x=2
ap dung dl pitago vao tam giac vuong BHI \(BI^2=14-x^2=14-4=10\Rightarrow BI=\sqrt{10}\)
. ............................HIC \(IC=\sqrt{40}\)
\(\Rightarrow BC=BI+IC=\sqrt{10}+\sqrt{40}\)
MA AI=\(AH+HI=8+2=10\)
\(\Rightarrow SABC=\frac{10.\left(\sqrt{10}+\sqrt{40}\right)}{2}=15\sqrt{10}\)
\frac{x}{x+8}+\frac{\frac{8}{\sqrt{14}}x}{\frac{8}{\sqrt{14}}x+\sqrt{14}}+\frac{\frac{8x}{\sqrt{44}}}{\frac{8x}{\sqrt{44}}+\sqrt{44}}=1x+8x+148x+14148x+448x+44448x=
A B C D E F H
a) Áp dụng định lí pitago.
Ta có: \(AB^2=AD^2+BD^2=BE^2+AE^2\)
\(HC^2=HD^2+DC^2=HE^2+EC^2\)
=> \(AB^2+HC^2=AD^2+BD^2+HD^2+DC^2\)
\(=\left(AD^2+DC^2\right)+\left(BD^2+HD^2\right)=AC^2+BH^2\) (1)
và \(AB^2+HC^2=BE^2+AE^2+HE^2+EC^2\)
\(=\left(BE^2+EC^2\right)+\left(AE^2+HE^2\right)=BC^2+AH^2\)(2)
Từ (1) , (2) Ta có: \(AB^2+HC^2=AC^2+HB^2=BC^2+HA^2\)
b) Ta có: \(S_{AHB}+S_{AHC}+S_{BHC}=S_{ABC}=S\)
\(AB.HC=AB\left(CF-FH\right)=AB.CF-AB.FH\)
\(=2S_{ABC}-2S_{AHB}=2S-2S_{ABH}\)
Tương tự: \(BC.HA=2S-2S_{BHC}\)
\(CA.HB=2S-2S_{AHC}\)
Cộng lại ta có:
\(AB.HC+BC.AH+CA.HB=6S-2\left(S_{AHB}+S_{AHC}+S_{BHC}\right)\)
\(=6S-2S=4S\)(đpcm)
A B C H D E F
Gọi D, E, F lần lượt là chân đường cao hạ từ A, B, C của tam giác ABC.
+) \(\Delta AHE~\Delta ACD\)( vì ^HAE =^CAD, ^HEA=^CDA )
=> \(\frac{HA}{CA}=\frac{EA}{AD}\)=> \(\frac{HA}{CA}.\frac{HB}{BC}=\frac{EA}{CA}.\frac{HB}{BC}=\frac{2.EA.HB}{2.CA.BC}=\frac{S_{\Delta AHB}}{S_{ABC}}\)(1)
+) \(\Delta CHD~\Delta CBF\)( vì ^DCH=^FCB, ^CDH=^CFB )
=> \(\frac{CH}{CB}=\frac{CD}{CF}\)=> \(\frac{CH}{CB}.\frac{AH}{AB}=\frac{CD.AH}{CF.AB}=\frac{S_{AHC}}{S_{ABC}}\)(2)
+) \(\Delta ABE~\Delta HBF\)
=> \(\frac{HB}{AB}=\frac{BF}{BE}\Rightarrow\frac{HB}{AB}.\frac{HC}{AC}=\frac{BF.HC}{BE.AC}=\frac{S_{BHC}}{S_{ABC}}\)(3)
Từ (1) ; (2) ; (3) => \(\frac{HA}{CA}.\frac{HB}{BC}+\frac{CH}{CB}.\frac{AH}{AB}+\frac{HB}{AB}.\frac{HC}{AC}=\frac{S_{ABE}}{S_{ABC}}+\frac{S_{ABE}}{S_{ABC}}+\frac{S_{ABE}}{S_{ABC}}=1\)
=> \(\frac{HA}{BC}.\frac{HB}{AC}+\frac{HB}{AC}.\frac{HC}{AB}+\frac{HC}{AB}.\frac{HA}{BC}=1\)
Đặt: \(\frac{HA}{BC}=x;\frac{HB}{AC}=y;\frac{HC}{AB}=z\); x, y, z>0
Ta có: \(xy+yz+zx=1\)
=> \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3\)
=> \(x+y+z\ge\sqrt{3}\)
"=" xảy ra khi và chỉ khi x=y=z
Vậy : \(\frac{HA}{BC}+\frac{HB}{AC}+\frac{HC}{AB}\ge\sqrt{3}\)
"=" xảy ra <=> \(\frac{HA}{BC}=\frac{HB}{AC}=\frac{HC}{AB}\)
a: Xét ΔABC vuông tại A có AH là đường cao
nên \(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\end{matrix}\right.\Leftrightarrow\dfrac{BH}{CH}=\dfrac{AB^2}{AC^2}\)
b: \(\dfrac{BE}{CF}=\dfrac{BH^2}{AB}:\dfrac{CH^2}{AC}=\dfrac{BH^2}{AB}\cdot\dfrac{AC}{CH^2}\)
\(=\dfrac{BH^2}{CH^2}\cdot\dfrac{AC}{AB}=\dfrac{AB^4}{AC^4}\cdot\dfrac{AC}{AB}=\dfrac{AB^3}{AC^3}\)
e: \(BE\cdot CF\cdot BC\)
\(=\dfrac{HB^2}{AB}\cdot\dfrac{HC^2}{AC}\cdot BC\)
\(=\dfrac{AH^4}{AB\cdot AC}\cdot BC=\dfrac{AH^4}{AH\cdot BC}\cdot BC=AH^3\)
\(=EF^3\)
gọi M;N;K là hình chiếu của A;B;C trên BC;AC;AB
A B C M K N H
Xét tan giác BHK và tam giác CHN là 2 tam giác đồng dạng (dễ dàng chứng minh) =>\(\frac{KH}{HB}=\frac{HN}{HC}< =>KH.HC=HB.HN\)
AB2=BN2+NA2=(BH+HN)2+HA2-HN2=BH2+2BH.HN+HA2=BH2+2CH.HK+HA2
AC2=AK2+KC2=(CH+HK)2+AH2-HK2=CH2+2CH.HK+AH2
BC2=CK2+KB2=(CH+HK)2+HB2-KH2=CH2+2CH.HK+HB2
=> AB2+HC2=AC2+HB2=BC2+HA2= CH2+2CH.HK+HB2+HA2
