Đáp án: $\{\dfrac14, \dfrac34\}$

 

Giải thích các bước giải:

Gọi $DE//AC, DF//AB, E\in AB, F\in AC$

$\to AEDF$ là hình bình hành

     $S_{AEDF}=\dfrac38S_{ABC}$

$\to S_{DBE}+S_{CDF}=S_{ABC}-\dfrac38S_{ABC}=\dfrac58S_{ABC}$

$\to \dfrac{S_{DBE}}{S_{ABC}}+\dfrac{S_{CDF}}{S_{ABC}}=\dfrac58$

$\to \dfrac{BD^2}{BC^2}+\dfrac{CD^2}{BC^2}=\dfrac58$

$\to \dfrac{BD^2+CD^2}{BC^2}=\dfrac58$

$\to \dfrac{(BC+CD)^2-2BD\cdot CD}{BC^2}=\dfrac58$

$\to \dfrac{BC^2-2BD\cdot CD}{BC^2}=\dfrac58$

$\to 1-\dfrac{2BD\cdot CD}{BC^2}=\dfrac58$

$\to \dfrac{2BD\cdot CD}{BC^2}=\dfrac38$

$\to \dfrac{BD\cdot CD}{BC^2}=\dfrac3{16}$

$\to \dfrac{BD\cdot (BC-DB)}{BC^2}=\dfrac3{16}$

$\to \dfrac{BD}{BC}-\dfrac{BD^2}{BC^2}=\dfrac3{16}$

Đặt $\dfrac{BD}{BC}=x, x>0$

$\to x-x^2=\dfrac3{16}$

$\to x\in\{\dfrac14, \dfrac34\}$