a || b
Construct a 4 angled figure ( could be Square, trapezoid, deltoid or whatever ) whose Surface will be equal to the surface of the pentagon.
I'll show you how to get whatever
Use shearing to get a quadrilateral:
$\displaystyle \Delta(ECD) \cong \Delta(ECD')~\implies~(ABCD')$ is a quadrilateral.
Use shearing again to get a triangle:
$\displaystyle (\overline{D'B}) \parallel (\overline{CC'})$ therefore: $\displaystyle \Delta(D'BC) \cong \Delta(D'BC')$
The area of the pentagon didn't change therefore the area of $\displaystyle \Delta(AC'D') = (ABCDE)$
Smart move.
But I how do we prove that the surface of rectangle EQE1 = the Surface of the Rectangle QDC
Ok maybe I am too stupid to understand this but as far as I know we need 3 elements in common for 2 Triangles to be equal as u stated ( Triangle ECD = Triangle ECD` ), as far as I can see those 2 triangles have only one thing in common and that is EC base.
The way question is worded does not prevent using an isoscelesCode:E E F D A D A C B C B
triangle on top of a square, then made into a rectangle.
EXAMPLE:
Square ABCD sides = 6, AE = DE = 5 ; so area ABCDE = 36 + 12 = 48.
Rectangle becomes a 8 by 6 (ADEF being 6 by 2), so area 48.
I don't think we have to do that.
Instead, we can do this:
$\displaystyle \mbox{Here the height, h, is the same for both}~\triangle {ECE_{1}}~\mbox{and}~\triangle {ECD}.$
As both triangles have the same base and the same height, they will also have the same area of
$\displaystyle \displaystyle{\frac{h(\mid{EC}\mid)}{2}}$