Area of Trapezoid

Let $CD=3x$ , $AB=2x$ , and the distance between the bases be $h$ . Then, $AE=EB=x$ .

Extend $\overline{DF}$ to hit $\overline{AB}$ at $Q$ . Then, $\triangle DFC\cong\triangle QFB$ by ASA. By this congruence, $EQ=4x$ . Also, $\triangle DPC\sim\triangle QPE$ by AA, and the height of $\triangle DPC$ is $\dfrac{3x}{3x+4x}h=\dfrac{3}{7}h$ . The area of $\triangle DPC$ is then $\dfrac{1}{2}\cdot 3x\cdot\dfrac{3}{7}h=\dfrac{9xh}{14}=45$ so $xh=70$ . The area of $ABCD$ is $\dfrac{3x+2x}{2}\cdot h=\dfrac{5}{2}xh=\boxed{175}$

Let $G$ be the midpoint of $AD,$ and draw $FG.$ Let $Q$ be the intersection point of lines $FG$ and $EC.$ FInally, let $AE = EB = x,$ giving $DC = 3x.$

Consider triangle $\triangle EBC.$ Since $AB, FG, CD$ are parallel and $F$ is the midpoint of $BC,$ we see that $QF = \dfrac{x}{2}.$ Now, consider triangles $\triangle DPC, \triangle QPF.$ Since the area of $\triangle DPC$ is $45$ and $DC = 3x,$ the altitude from $P$ to side $DC$ must be $\dfrac{30}{x}.$ Furthermore, since $QF \parallel DC$ and $DC = 6QF,$ triangles $\triangle DPC, \triangle QPF$ are similar with a ratio of $\, 6 : 1.$ Then, the altitude from $P$ to side $QF$ must be $\dfrac{1}{6} \cdot \dfrac{30}{x} = \dfrac{5}{x}.$ Thus the distance between parallel lines $FG$ and $CD$ is $\dfrac{30}{x} + \dfrac{5}{x} = \dfrac{35}{x}.$

But since $FG$ is the midsegment of trapezoid $ABCD,$ the altitude of the trapezoid must be $2 \cdot \dfrac{35}{x} = \dfrac{70}{x}$ ; therefore, the area of the trapezoid is $\dfrac{2x+3x}{2} \cdot \dfrac{70}{x} = \boxed{175}.$

{ABCD} means the area of ABCD. DP intersects ray AB at K. We have: BK=DC; EB=1/3 DC. Hence EB+BK= 4/3DC => EK=4/3 DC => EP=4/3 PC => {EPD}=4/3{DPC}=4/3.45=60 => {EDC}=60+45=105. We have: ({AED}+{EBC}) / {EDC}= (AE+EB) / DC= 2/3 => {AED}+{EBC}=105. 2/3=70. So, {ABCD}= {AED}+{EBC}+{EDC}= 70+105=175