[tex]In~triunghiul~ADE~observam~ca~QG~||~AD~si~QC= \frac{BC}{2}= \frac{AD}{2}. \\ \\ Rezulta~ca~[QC]~este ~linie~mijlocie~in~ \Delta ADE \Rightarrow~C-mijlocul~lui \\ \\ \big [DE \big].[/tex]
[tex]Deoarece~P~si~Q~sunt~mijloacele~laturilor~[AD]~si~[BC],~rezulta \\ \\ ca~PQ~||~CD~si~PQ=CD~(ABQP-dreptunghi) ~deci~PQ \perp BC \Leftrightarrow \\ \\ \Leftrightarrow PQ \perp CQ. \\ \\ Deoarece~PQ~||~CE,~rezulta~ca~d(E,PQ)=d(C,PQ)=CQ= \frac{BC}{2}= \\ \\ =6~(cm). \\ \\ A_{PQE}= \frac{PQ \cdot d(E,PQ)}{2}= \frac{PQ \cdot d(C,PQ)}{2}= \frac{12 \cdot 6}{2}=36~(cm^2). [/tex]
