[Toán 9]Hình học

I

i_love_cat

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C

congchuaanhsang

Kẻ HE//AB (E thuộc KM)
Theo Ta-lét ta có: $\dfrac{KH}{AK}$=$\dfrac{HE}{AM}$=$\dfrac{HE}{BM}$=$\dfrac{HD}{BD}$

AD là phân giác $\hat{BAH}$\Rightarrow$\dfrac{HD}{BD}$=$\dfrac{AH}{AB}$

\Rightarrow$\dfrac{HK}{AK}$=$\dfrac{AH}{AB}$ (1)

Mặt khác $\hat{DAC}$=$\hat{DAH}$+$\hat{HAC}$=$\hat{BAD}$+$\hat{ABD}$=$\hat{ADC}$
\Rightarrow$\Delta$ADC cân ở C\RightarrowAC=CD
$\Delta$ABC vuông ở A\Rightarrow$AC^2$=BC.CH\LeftrightarrowAC.CD=BC.CH

\Leftrightarrow$\dfrac{AC}{BC}$=$\dfrac{CH}{CD}$
$\Delta$ABH $\sim$ $\Delta$CBA (g.g)

\Rightarrow$\dfrac{AH}{AB}$=$\dfrac{AC}{BC}$ = $\dfrac{CH}{CD}$ (2)

Từ (1) và (2)\Rightarrow$\dfrac{HK}{AK}$=$\dfrac{CH}{CD}$ \RightarrowCK//AD
 
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