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
Sorry it's been a long time. I was away in Canada when you posted. Since my return, I've had lots to do, so your problem got put to one side. But now I have a complete method. Hope you like it.
In this diagram I show the line AB and the angle D.
Extend AB and also one of the lines from D until they cross. Construct a circle through this point, B and D using perpendicular bisectors. Extend the other arm of D to cut the circle. Using the equal angle property of a circle this point, joined to B will be the same size as D.
(I have also put on a random point E. Later I adjust its position on AB extended to get the right size for BE.)
In this next diagram I have hidden most of the construction from above, to keep the diagram simple.
I have moved E off to one side to keep it out of the way. Now I show triangle C. By use of parallel lines I have translated C down so that one corner is on BE. This new position for the triangle is shown in yellow. Using that point on BE as a centre I have made two circles. My next aim is to rotate the yellow triangle until one side lies on BE. You can see I have marked a target point on BE. That's where this triangle will rotate to.
I have added some letters so you know which points I am talking about. My aim is to rotate y onto x. By extending the other side of the yellow triangle to point z, I am ready to proceed. Use a compass to set up the distance yz. With that radius and x as centre, step off around the circle to find point w. By drawing out from the centre of the circle I can reproduce an exact copy of triangle C, but now lying along BE.
It should not be necessary to make any more diagrams as the rest is very straight forward.
Use parallels to 'slide' the triangle left until x lies on position B. E is now the other end of that side of the triangle. Bisect the other side to fix point G. Draw GF parallel to BE.
The parallelogram BEFG has angle D and area C as required.
footnote. I seem to have accidentally made the angle in triangle C look very much the same as D. It doesn't change the method though. By bisecting a side of the triangle we can find its height, and any other triangle with the same base and height will have the same area. A line parallel to BE at that height distance from BE will fix G.
Last edited by bob bundy (2014-08-02 01:51:25)
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