Question 116
Let $A$, $C$ be interior points of square $XOBD$, such that $\angle AXC = \angle ABC = 45^\circ$. Prove that,
$$
S_{\triangle AXO} + S_{\triangle ABC} + S_{\triangle CXD} = S_{\triangle ACX} + S_{\triangle AOB} + S_{\triangle CBD}.
$$
$$
S_{\triangle AXO} + S_{\triangle ABC} + S_{\triangle CXD} = S_{\triangle ACX} + S_{\triangle AOB} + S_{\triangle CBD}.
$$