AB is a line
D is given angle
C is given triangle
We have to construct a parallelogram which has area equal to triangle C, with one side as AB & one angle equal to D
Now the way it's been proven is
A parallelogram BEFG is constructed with BE in line with AB & FG below it &angle D is equal to angle EBG & area of triangle C equal to area of this parallelogram(I know how to construct a parallelogram having area equal to a given triangle if they are on the same line but I don't know how to construct when they , as is shown in the figure which I couldn't upload, the trangle C is shown in a separate corner & not even near BEFG, in that case how do I construct the parallelogram having the same area is my question)
Coming back to the proof
Extend FG upto a certain point H & join AH such that AH is parallel to EF
Join H & B.
Now since HF is on 2 parallels AH & EF then angle AHF + angle HFE = 180 degrees
Also angle BHG + angle GFE(or HFE) is < 180 degrees ( as they intersect in that direction)
Since they must meet we extend HB & FE upwards where they meet at say K.
We draw KL || EA
Now extend HA & GB to meet KL at points L & say M respectively
Now the whole LKFH is a ||gram with HK as diagonal
Also inside it MKEB & ABGH are ||grams & LMBA & BEFG are complementary ||grams (on opposite side of the diagonal)
So bya previous proof complements on either side of a diagonal are equal in are
So Area LMBA = Area BEFG = Area triangle C
Angle GBE = Angle ABM = Angle D (since vertically opposite angles, if I am right)
So we have a parallelogram LMBA with Ab as its side, area same as C & an angle equal to D.
The remaining parts I have understood because there are proofs for all those etc, but only the step where I have noted is what I don't understand.
Thanking you again
* Earlier I had posted this for which the diagram given by bob was correct but I wherein he suggested that he could construct it if one of the side were parallel to the line but I still don't get how. Please help