Proof Concerning Central Line \(X_5X_6\) of Triangle

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Hello everyone. Today we will prove theorem about central line \(X_5X_6\) of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let \(ABCD\) be a golden rectangle where \( \frac{AB}{BC} = \phi \), and construct the square \( BCQP \) inside it. Reflect \( P \) over \( D \) to obtain \( E \). Then, the line \( EB \) coincides with the central line \( X_5X_6 \) of triangle \( ABP \). Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: \( A = (0,0) \), \( B = (\phi x, 0) \), \( C = (\phi x, x) \), \( D = (0, x) \). The square \( BCPQ \) ensures \( P = (\phi x, 2x) \). The reflection of \( P \) across \( D \) is \( E = (-\phi x, 2x) \). 2. Compute the Nine-Point Center \( X_5 \) The nine-point center \( X_5 \) is the circumcenter of the medial triangle, which consists of the midpoints: \[ M_1 = \left(\frac{0 + \phi x}{2}, 0\right) = \left(\frac{\phi x}{2}, 0\right), \] \[ M_2 = \left(\f...
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Proof that the square root of 2 is an irrational number

 Hello everyone. Today we will prove that \(\sqrt{2} \) is an irrational number. We will use the method of contradiction as a proof method. This proof was first given by the Greek philosopher Aristotle. 

Theorem: \(\sqrt{2} \) is an irrational number. 

Proof: 

Suppose that \(\sqrt {2} \) is a rational number. Then \(\sqrt {2} \) can be written in the form \(\frac{p}{q} \) where \(p \) and \(q \) are coprime integers such that \(q \neq 0 \). Note that since \(\frac{p}{q} \) is an irreducible fraction \(p \) and \(q \) cannot be even at the same time, otherwise the fraction would not be irreducible. 

From the equation \(\sqrt{2} = \frac{p}{q} \) it follows that \(2 = \frac{p^2}{q^2} \) or \(p^2 = 2q^2 \). So \(p^2 \) is an even number from which it follows that \(p \) is also an even number so we can write the number \(p \) in the form \(p = 2k \) , where \(k \) is an integer . 

So, we have that \((2k)^2 = 2q^2 \) i.e. \(4k^2 = 2q^2 \), or \(q^2 = 2k^2 \). From the last equation we conclude that \(q^2 \) is an even number, so \(q \) is an even number too. 

We have shown that \(p \) and \(q \) are even numbers, which contradicts the assumption that \(\frac{p}{q} \) is an irreducible fraction. Thus, the initial assumption that \(\sqrt{2} \) is a rational number is incorrect, which means that \(\sqrt{2} \) is an irrational number. 

\(\blacksquare \) 

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