四角形の対角線と面積

$\triangle\text{ABP}$、$\triangle\text{BCP}$、$\triangle\text{CDP}$、$\triangle\text{DAP}$ の面積を、それぞれ $S_1$、$S_2$、$S_3$、$S_4$ とおくと \begin{align} S_1=&\dfrac{1}{2}\text{AP}\cdot\text{BP}\sin\theta\\ &\blacktriangleleft 三角形の面積参照\\ S_2=&\dfrac{1}{2}\text{BP}\cdot\text{CP}\sin(180^\circ-\theta)\\ =&\dfrac{1}{2}\text{BP}\cdot\text{CP}\sin\theta\\ &\blacktriangleleft 180^\circ-\thetaの三角比\\ S_3=&\dfrac{1}{2}\text{CP}\cdot\text{DP}\sin\theta\\ S_4=&\dfrac{1}{2}\text{DP}\cdot\text{AP}\sin(180^\circ-\theta)\\ =&\dfrac{1}{2}\text{DP}\cdot\text{AP}\sin\theta \end{align} となるから \begin{align} S=&S_1+S_2+S_3+S_4\\ =&\dfrac{1}{2}(\text{AP}\cdot\text{BP}+\text{BP}\cdot\text{CP}\\ &\qquad+\text{CP}\cdot\text{DP}+\text{DP}\cdot\text{AP})\sin\theta\\ =&\dfrac{1}{2}\{\text{BP}\cdot(\text{AP}+\text{CP})\\ &\qquad+\text{DP}\cdot(\text{CP}+\text{AP})\}\sin\theta\\ =&\dfrac{1}{2}\left(\text{AP}+\text{CP}\right)\left(\text{BP}+\text{DP}\right)\sin\theta\\ =&\dfrac{1}{2}\text{AC}\cdot\text{BD}\sin\theta=\frac{1}{2}ab\sin\theta \end{align}