We had to draw a parallelogram with a given line as its side & area equal
to a given triangle & also that one of the angles in the parallelogram
would be equal to a given angle
I understood everything in the proof except one that is that the given line
say AB was extended to E & that the parallelogram BEFG was constructed with
area equal to the given triangle, but the base of the triangle & the line
BE were not on the same line they were different, I know how to construct a
parallelogram with one of its sides in line with the base of a given
triangle & with their areas being equal but I don`t understand how to
construct it when the side of the parallelogram is not inline with the base
of the triangle.
Please help me
Please would you provide a diagram for this. I'm getting lost with which line is which?
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sir, How can I upload the image?
Last edited by Sven (2014-06-10 17:38:55)
I use imgur.com
Or maybe you could just describe it in words like this for example:
AB is a horizontal line. C is a point somewhere above the line.
ABC is a triangle.
D is the midpoint of AC and E is the midpoint of BC.
Join DE. DE is parallel to AB and the perpendicular height of DE above AB is half the perpendicular height of ABC.
Draw a line BF parallel to AD and cutting DE produced at F.
Then ABFD is a parallelogram with the same area as ABC.
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
Here's my diagram:
I can do the construction easily if one of the sides of C were parallel to AB and similarly the angle D. It gets tougher if they are like I've shown. Once rotated to make the parallel property, I'm ok but rotating is not a 'pencil and ruler' construction as far as I know.
The diagram is absolutely correct now I just want to know how to do the construction if you could tell me.