Toán hình học lớp 8

H

heoxinh9kc

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T

tsukishizuku

[TEX][/TEX] Cho tam giác ABC nhọn.Ba đường cao AA1,BB1,CC1 cắt nhau tại H.Chứng minh: $\dfrac{A_1H}{A_1A} + \dfrac{B_1H}{B_1B} + \dfrac{C_1H}{C_1C}=1$(Giúp mình sớm nhé).

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Ta có:
$2.S_(ABC)=AA_1.BC=BB_1.AC=CC_1.AB$
Ta có:
$2.S_(AHC)=HB_1.AC;2.S_(BHC)=HA_1.BC;2.S_(AHB)=HC_1.AB$
Ta có:
$\dfrac{A_1H}{A_1A} + \dfrac{B_1H}{B_1B} + \dfrac{C_1H}{C_1C}$
$=\dfrac{A_1H.BC}{A_1A.BC} + \dfrac{B_1H.AC}{B_1B.AC} + \dfrac{C_1H.AB}{C_1C.AB}$
$=\frac{2.S_(BHC)}{2.S_(ABC)} + \frac{2.S_(AHC)}{2.S_(ABC)} + \frac{2.S_(AHB)}{2.S_(ABC)}$
$=\frac{2.S_(BHC) + 2.S_(AHC) +2.S_(AHB)}{2.S_(ABC)}$
$=\frac{2.(S_(BHC) + S_(AHC) +S_(AHB))}{2.S_(ABC)}$
Mà do tam giác ABC nhọn \Rightarrow $S_(BHC) + S_(AHC) +S_(AHB)=S_(ABC)$
\Rightarrow$=\frac{2.(S_(BHC) + S_(AHC) +S_(AHB))}{2.S_(ABC)}$=$\frac{2.S_(ABC)}{2.S_(ABC)}=1$
(đpcm)
 
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