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24 Oct 2011 00:12:02 IST

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Prove Euler's Quadrilateral theorem by the method of vectors.

Mathematics

Let ABCD be any Quadrilateral where P and Q are the midpoints of the diagonals. Prove that

AB2+BC2+CD2+DA2=AC2+BD2+4PQ2.

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Sumit Bhattacharjee

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Joined: 18 Oct 2011

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24 Oct 2011 00:37:28 IST

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Draw any quadrilateral ABCD. Join the diagonals AC and BD. Let P and Q are the respective midpoints of the diagonals. We have to prove that AB²+BC²+ AB²+BC²+CA²+DA²=AC²+BD²+4PQ².

Proof: By construction we can see that AP+PQ+QB=AB, or ½(AC-BD)+PQ=AB……(i)

CP+PQ+QD=CD, or ½(BD-AC)+PQ=CD, or ½(AC-BD)-PQ=CD……..(ii)

BQ+QP+PC=BC, or ½(AC+BD)-PQ=BC………(iii)

AP+PQ+QD=AD, or ½(AC+BD)+PQ=DA…….(iv)

Now, let PQ makes an angle α with (AC-BD), so that,

{1/2(AC-BD)+PQ}²=1/4(AC-BD)²+PQ²+2.1/2(AC-BD).PQ cos α ….(v) [AC

{ 1/2(AC-BD)-PQ}² must be equal to 1/4(AC-BD)²+PQ²-2.1/2(AC-BD).PQ cos α ….(vi)

Similarly, if { 1/2(AC+BD)+PQ}²=1/4(AC+BD)²+PQ²+2.1/2(AC+BD).PQ cos β….(vii) [letting the angle between (AC+BD) and PQ be β]

Then, { 1/2(AC+BD)-PQ}²=1/4(AC+BD)²+PQ²-2.1/2(AC+BD).PQ cos β ….(viii).

Adding the equations with (v), (vi), (vii) and (viii) together with (i), (ii), (iii) and (iv), we get

AB²+BC²+CD²+DA²=AC²+BD²+4PQ²

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