ЛЕМА и ЗАКЛУЧОК кои не се премногу НЕКОРИСНИ // A LEMMA and a COROLLARY that are not too USELESS

          We denote in these lines with $\textbf{S}_{ABC}$ the area of ​​triangle $ABC$.

          LEMMA  Let $ABCD$ be a convex quadrilateral and $O$ be the point of

                           intersection of the diagonals $AC,\;BD.$ Then

$$\textbf{S}_{AOB} \cdot \textbf{S}_{COD}= \textbf{S}_{AOD} \cdot \textbf{S}_{BOC} \tag{1}$$

                          holds.

         Proof of Lemma {it would work immediately by applying the formula for the area of ​​a triangle $\textbf{S}_{AOB}=OA \cdot OB \cdot \sin \widehat{AOB}/2$ etc. and the equality of the sines of the supplementary angles. I prefer a demonstration that would make even a parrot envious.}

          Let ${\color{Brown} {AE}},\;{\color{Brown}{CF}} \perp BD$. 

We have the formulas

$$2\textbf{S}_{AOB}={\color{Yellow}{OB}} \cdot {\color{Brown}{AE}},\;\;\;\;2\textbf{S}_{BOC}={\color{Blue}{OB}} \cdot {\color{Brown}{CF}} \tag{2}$$

$$2\textbf{S}_{COD}={\color{Green}{OD}} \cdot {\color{Brown}{CF}},\;\;\;\;2\textbf{S}_{AOD}={\color{Red}{OD}} \cdot {\color{Brown}{AE}} \tag{3}$$

Then

$4\textbf{S}_{AOB}\cdot \textbf{S}_{COD}={\color{Yellow}{OB}} \cdot {\color{Brown}{AE}} \cdot {\color{Green}{OD}} \cdot {\color{Brown}{CF}}={\color{Blue}{OB}} \cdot {\color{Brown}{CF}} \cdot {\color{Red}{OD}} \cdot {\color{Brown}{AE}}=4\textbf{S}_{BOC} \cdot \textbf{S}_{AOD}.$

The Lemma is proven.

$\square$(QED Lemma)

          COROLLARY  In the trapezoid $ABCD,\;\; AB \parallel CD$ we have

$$\textbf{S}_{AOD}^2=\textbf{S}_{BOC}^2=\textbf{S}_{AOB} \cdot \textbf{S}_{COD} \tag{4}$$

          Indeed, let $AE,\;BF \perp CD$. We have $AE=BF$, so

$2\textbf{S}_{ACD}=AE \cdot CD=BF \cdot CD=2 \textbf{S}_{BCD}\;\;\Rightarrow\;\textbf{S}_{ACD}-\textbf{S}_{COD}=\textbf{S}_{BCD}-\textbf{S}_{COD}\Rightarrow$

$$\Rightarrow\;\;\textbf{S}_{AOD}=\textbf{S}_{BOC}$$

and (4) results from (1).

$\blacksquare$