# Understanding Ceva's Theorem In Ceva's Theorem, I understand that $$\dfrac{A_{\triangle PXB}}{A_{\triangle PXC}}=\dfrac{BX}{CX}=\dfrac{A_{\triangle BXA}}{A_{\triangle CXA}}$$.

I would like clarification in understanding the following step which states:

$$\frac{A_{\triangle APB}}{A_{\triangle APC}}=\frac{A_{\triangle AXB} - A_{\triangle PXB}}{A_{\triangle AXC}-A_{\triangle PXC}}=\frac{BX}{CX}$$

How does the subtraction of the two areas make it so that the new triangles, even though they do not share those sides, are still proportional to $$\frac{BX}{CX}$$?

## 1 Answer

$$A_{\triangle AXB}: A_{\triangle AXC}=BX:CX\Rightarrow A_{\triangle AXB}=\frac{BX}{CX}A_{\triangle AXC}$$

$$A_{\triangle PXB}: A_{\triangle PXC}=BX:CX\Rightarrow A_{\triangle PXB}=\frac{BX}{CX}A_{\triangle PXC}$$

Hence $$\frac{A_{\triangle A X B} -A _{\triangle P X B}}{A _{\triangle A X C}-A_{ \triangle P X C}}=\frac{\frac{BX}{CX}A_{\triangle AXC}-\frac{BX}{CX}A_{\triangle PXC}}{A _{\triangle A X C}-A_{ \triangle P X C}}=\frac{BX}{CX}$$