$\dfrac{SA}{SA'}.S_{BCD}+\dfrac{SC}{SC'}.S_{ABD}=\dfrac{SB}{SB'}.S_{ACD}+\dfrac{SD}{SD'}.S_{ABC}$

Bài toán:
Cho hình chóp $S.ABCD$ có đáy là tứ giác lồi. Mặt phẳng $(P)$ cắt các cạnh $SA,SB,SC, SD$ lần lượt tại $A', B', C', D'$. Chứng minh rằng:
$$\dfrac{SA}{SA'}.S_{BCD}+\dfrac{SC}{SC'}.S_{ABD}=\dfrac{SB}{SB'}.S_{ACD}+\dfrac{SD}{SD'}.S_{ABC}$$ .



Lời giải:

$$\begin{align*} \dfrac{V_{S.A'B'C'}}{V_{S.ABC}}&=\dfrac{SA'}{SA}.\dfrac{SB'}{SB}.\dfrac{SC'}{SC}\\ \Leftrightarrow V_{S.A'B'C'}&=\dfrac{1}{3}h.S_{ABC}.\dfrac{SA'}{SA}.\dfrac{SB'}{SB}.\dfrac{SC'}{SC} \end{align*}$$

Tương tự ta có:

$$V_{S.A'D'C'}=\dfrac{1}{3}h.S_{ACD}.\dfrac{SA'}{SA}.\dfrac{SD'}{SD}.\dfrac{SC'}{SC}$$

$$\begin{align*} \Rightarrow V_{S.A'B'C'D'}&=V_{S.A'B'C'}+V_{S.A'D'C'}\\ &=\dfrac{1}{3}h.\dfrac{SA'}{SA}.\dfrac{SC'}{SC}\left ( \dfrac{SB'}{SB}.S_{ABC}+\dfrac{SD'}{SD}.S_{ACD} \right ) \end{align*}$$

Tương tự:

$$V_{S.A'B'C'D'}=\dfrac{1}{3}h.\dfrac{SB'}{SB}.\dfrac{SD'}{SD}\left ( \dfrac{SA'}{SA}.S_{ABD}+\dfrac{SC'}{SC}.S_{BCD} \right )$$

$$\Rightarrow \dfrac{1}{3}h.\dfrac{SA'}{SA}.\dfrac{SC'}{SC}\left ( \dfrac{SB'}{SB}.S_{ABC}+\dfrac{SD'}{SD}.S_{ACD} \right )=\dfrac{1}{3}h.\dfrac{SB'}{SB}.\dfrac{SD'}{SD}\left ( \dfrac{SA'}{SA}.S_{ABD}+\dfrac{SC'}{SC}.S_{BCD} \right )$$

$$\Leftrightarrow \dfrac{SA}{SA'}.S_{BCD}+\dfrac{SC}{SC'}.S_{ABD}=\dfrac{SB}{SB'}.S_{ACD}+\dfrac{SD}{SD'}.S_{ABC}. \blacksquare$$