Giải thích các bước giải:
Qua $O$ kẻ $BD//A_1C_1$
$\to \dfrac{SA_1}{SB}=\dfrac{SC_1}{SD}=\dfrac{SO_1}{SO}=\dfrac29$
$\to SC_1=\dfrac29SD$
Mà $\dfrac{SC_1}{SC}=\dfrac15\to SC=5SC_1=\dfrac{10}9SD$
$\to \dfrac{CS}{SD}=\dfrac{10}{9}$
$\to \dfrac{SC-SD}{SD}=\dfrac{10-9}{9}$
$\to\dfrac{CD}{DS}=\dfrac{1}{9}$
Vì $O$ là trung điểm $AC\to OA=OC\to \dfrac{OA}{OC}=1$
Ta có $B,O,D$ thẳng hàng
$\to \dfrac{BS}{BA}\cdot\dfrac{OA}{OC}\cdot\dfrac{DC}{DS}=1$
$\to \dfrac{BS}{BA}\cdot 1\cdot\dfrac{1}{9}=1$
$\to \dfrac{BS}{BA}=9$
$\to BS=9BA$
$\to BS-BA=8BA$
$\to SA=8AB=\dfrac89SB\to SB=\dfrac98SA$
Lại có:
$\dfrac{SA_1}{SB}=\dfrac29$
$\to SA_1=\dfrac29SB$
$\to SA_1=\dfrac29\cdot \dfrac98SA$
$\to SA_1=\dfrac14SA$
$\to \dfrac{SA_1}{SA}=\dfrac14$
