Question 1006481: Prove that in a parallelogram the sum of the squares of the diagonals equals the sum of the squares of the four sides
Found 2 solutions by ikleyn, jim_thompson5910:
Answer by ikleyn(52781)   (Show Source):
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website! Claim: In a parallelogram the sum of the squares of the diagonals equals the sum of the squares of the four sides
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I'm going to use coordinate geometry and the distance formula to prove the claim
Let's define a parallelogram ABCD (AB || CD and AC || BD) such that the four points are
A = (0,0)
B = (p,0)
C = (p+q,r)
D = (q,r)
where p,q, and r are constants (fixed numbers)
Let's make p,q,r positive numbers
Let's also define these distances
m = d(A,B) = d(C,D)
n = d(B,C) = d(A,D)
s = d(A,C)
t = d(B,D)
the notation d(A,B) means "the distance between points A and B"
The distance formula will be used here
d(P,Q) = sqrt((x1-x2)^2 + (y1-y2)^2)
where P = (x1,y1) and Q = (x2,y2)
Let's calculate the lengths of the sides
Distance from A to B (ie length of side AB)
d(A,B) = sqrt((x1-x2)^2 + (y1-y2)^2)
d(A,B) = sqrt((0-p)^2 + (0-0)^2)
d(A,B) = sqrt((-p)^2 + (0)^2)
d(A,B) = sqrt(p^2+0)
d(A,B) = sqrt(p^2)
d(A,B) = p
Distance from B to C (ie length of side BC)
d(B,C) = sqrt((x1-x2)^2 + (y1-y2)^2)
d(B,C) = sqrt((p-(p+q))^2 + (0-r)^2)
d(B,C) = sqrt((p-p-q)^2 + (-r)^2)
d(B,C) = sqrt((-q)^2 + (-r)^2)
d(B,C) = sqrt(q^2+r^2)
Distance from C to D (ie length of side CD)
d(C,D) = sqrt((x1-x2)^2 + (y1-y2)^2)
d(C,D) = sqrt(((p+q)-q)^2 + (r-r)^2)
d(C,D) = sqrt((p)^2 + (0)^2)
d(C,D) = sqrt(p^2)
d(C,D) = p
Distance from A to D (ie length of side AD)
d(A,D) = sqrt((x1-x2)^2 + (y1-y2)^2)
d(A,D) = sqrt((0-q)^2 + (0-r)^2)
d(A,D) = sqrt((-q)^2 + (-r)^2)
d(A,D) = sqrt(q^2+r^2)
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Let's calculate the lengths of the diagonals
A = (0,0)
B = (p,0)
C = (p+q,r)
D = (q,r)
Distance from A to C (ie length of diagonal AC)
d(A,C) = sqrt((x1-x2)^2 + (y1-y2)^2)
d(A,C) = sqrt((0-(p+q))^2 + (0-r)^2)
d(A,C) = sqrt((-(p+q))^2 + (-r)^2)
d(A,C) = sqrt((p+q)^2 + r^2)
d(A,C) = sqrt(p^2+2pq+q^2 + r^2)
Distance from B to D (ie length of diagonal BD)
d(B,D) = sqrt((x1-x2)^2 + (y1-y2)^2)
d(B,D) = sqrt((p-q)^2 + (0-r)^2)
d(B,D) = sqrt((p-q)^2 + (-r)^2)
d(B,D) = sqrt(p^2-2pq+q^2 + r^2)
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Summary so far:
sides
m = d(A,B) = p
m = d(C,D) = p
n = d(B,C) = sqrt(q^2+r^2)
n = d(A,D) = sqrt(q^2+r^2)
diagonals
s = d(A,C) = sqrt(p^2+2pq+q^2 + r^2)
t = d(B,D) = sqrt(p^2-2pq+q^2 + r^2)
Notice we get this if we square both sides of each equation (above)
m^2 = p^2
n^2 = q^2+r^2
s^2 = p^2+2pq+q^2 + r^2
t^2 = p^2-2pq+q^2 + r^2
Squaring was done to get rid of the square roots.
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The claim is that "the sum of the squares of the diagonals equals the sum of the squares of the four sides"
In our parallelogram, the sides are m & n. The diagonals are s & t.
The translation of the claim, into an equation, is this
m^2 + n^2 + m^2 + n^2 = s^2 + t^2
which simplifies to
2m^2 + 2n^2 = s^2 + t^2
If we can prove the equation 2m^2 + 2n^2 = s^2 + t^2 is true, then we have proven the claim to be true.
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The good news is that at this point, that much is trivial. Perform substitutions to get...
2m^2 + 2n^2 = s^2 + t^2
2p^2 + 2(q^2+r^2) = p^2+2pq+q^2 + r^2 + p^2-2pq+q^2 + r^2
2p^2 + 2q^2 + 2r^2 = p^2+2pq+q^2 + r^2 + p^2-2pq+q^2 + r^2
2p^2 + 2q^2 + 2r^2 = p^2+q^2 + r^2 + p^2+q^2 + r^2
2p^2 + 2q^2 + 2r^2 = 2p^2+2q^2 + 2r^2 ... equation is true
Since the last equation is true for all values of p,q,r, this means 2m^2 + 2n^2 = s^2 + t^2 is true as well
So the claim has been verified and this ends the proof.
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