Survival of the fastest!?

Question

Ok Math fans, here's another challenge for you, a question that came up to my mind.
The first one to answer this correctly, with providing all the procedures of doing it, wins 10 whole points.
So here is the question:
Prove that { the difference between the two diagonals} times {the sum of the two diagonals} of any parallelogram equals to (4ab*cosx).
That is, prove that:
(d+f)(d-f) = (4ab)(cosx)
where d is the long diagonal, and f is the short one.
a and b are the two different sides of the parallelogram, and x is the small angle between those two sides. Assume that the other angle is (180 -x).
After providing the proof, give an example.
So, as said, "survival of the fastest!" Nice quote.

Answer 1

Easy..Take 2 triangles within the parallelogram with 2 angles: x and (180-x)

say d is the diagonal directly opposite that of x and so forth.

Use cosine rule for x :

d^2 = a^2+b^2-2abcosx--------------Eqn 1

Similarly 4 (180-x):

f^2 = b^2+a^2-2abcos(180-x)

f^2 = b^2+a^2+2abcosx---------------Eqn 2

Simultaneous:

Eqn 1-2: d^2 - f^2 = -4abcosx
(d-f)(d+f) = -4abcosx and since it's negative it means in my diagram-yes I can't visualise dis in my head- I should reverse my angle so x is obtuse and (180-x) is acute.

Also u must say sum and difference of the LENGTHS of the diagonals. You cannot operate mere lines.

Answer 2

This is easily seen with the law of cosines:

We have a parallelogram ABCD, where AC=d and BD=f.

Consider the triangle ABC:

The angle B=π-x.
The law of cosines gives:
d^2=a^2+b^2-2ab•cos(π-x) = a^2+b^2+2ab•cos(x).

Now consider the triangle ABD:

The angle A=x.
The law of cosines gives:
f^2=a^2+b^2-2ab•cos(x).

Therefore (d+f)(d-f) = d^2-f^2 = (a^2+b^2+2ab•cos(x)) - (a^2+b^2-2ab•cos(x)) = 4ab•cos(x).

Here is a stupid example:

Think of a square with diagonals d and f. Since it is a square d=f, therefore d-f=0 and (d+f)(d-f)=0. Also since it is a square the angle at all of the vertexes is π/2, and cos(π/2) = 0, thus 4ab•cos(x) = 0 and it works.

Or for something non-trivial (and more concrete): Let ABCD (A and B on the bottom, C and D on the top, it goes ABCD in counterclockwise order) have sides of length 3 (AB and CD) and 3√2 (AD and BC), with angle A = π/4.

Drawing this, you can see that BD is a vertical line, thus makes a right angle to AB. Using the Pythagorean theorem, f^2=(3√2)^2-3^2=9, thus f=3.

This gives more information though, it tells us that point C is 3+3 units to the right of A, and 3 units above A, Thus d^2=(3+3)^2+3^2= 45.

Therefore (d+f)(d-f) = d^2-f^2 = 45-9 = 36.

And cos(x) = cos(π/4) = √2/2, thus 4ab•cos(x) = 4•3•3√2•√2/2 = 36.

Answer 3

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Answer 4

using law of cosines:

f^2 = a^2 + b^2 - 2abcos(x)
d^2 = a^2 + b^2 - 2abcos(180-x) = a^2 + b^2 + 2abcos(x)

add:
2abcos(x) = a^2 + b^2 - f^2 and
2abcos(x) = d^2 - a^2 - b^2

so:
4abcos(x) = d^2 - f^2 = (d+f)(d-f)