Đáp án: $S_{KBC}=\dfrac35S$
Giải thích các bước giải:
Ta có:
$AD=\dfrac14AB$
$\to \dfrac{AD}{AB}=\dfrac14$
$\to\dfrac{AB-AD}{AB}=\dfrac{4-1}{4}$
$\to\dfrac{BD}{AD}=\dfrac34$
Ta có:
$AD=\dfrac14AB, AE=\dfrac14AC$
$\to \dfrac{AD}{AB}=\dfrac{AE}{AC}(=\dfrac14)$
$\to DE//BC$
$\to\dfrac{KD}{KC}=\dfrac{DE}{BC}=\dfrac{AD}{AB}=\dfrac14$
$\to \dfrac{KD+KC}{KC}=\dfrac{1+4}{4}$
$\to \dfrac{CD}{KC}=\dfrac54$
$\to \dfrac{KC}{CD}=\dfrac45$
$\to \dfrac{S_{KBC}}{S_{BCD}}=\dfrac45$
$\to S_{KBC}=\dfrac45S_{BCD}=\dfrac45\cdot \dfrac{BC}{AB}\cdot S_{ABC}$
$\to S_{KBC}=\dfrac45S_{BCD}=\dfrac45\cdot \dfrac{3}{4}\cdot S$
$\to S_{KBC}=\dfrac35S$
