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 of the formula for the sum of the first n natural numbers

Hello everyone. Today we will prove the formula for the sum of the first \(n \) natural numbers. We will use the method of mathematical induction as a proof method. It is believed that the Pythagoreans also knew this formula. 

Theorem: For any natural number \(n \) the following formula holds: \[\displaystyle \sum_{k = 1} ^ nk = \frac {n (n + 1)} {2} \] Proof: 

1. Base case (n = 1) \[\displaystyle \sum_{k = 1} ^ {1} k = \frac {1 \cdot (1 + 1)} {2} \] \[1 = \frac {1 \cdot (2)} {2} \] \[1 = \frac {2} {2} \] \[1 = 1 \] 

2. Induction hypothesis (n = m) 

Suppose that: \[\displaystyle \sum_{ k = 1} ^ mk = \frac {m (m + 1)} {2} \] 

3. Inductive step (n = m + 1) 

Using the assumption from the second step, we will prove that: \[\displaystyle \sum_{k = 1} ^ {m + 1} k = \frac {(m + 1) (m + 2)} {2} \] So, \[\displaystyle \sum_{k = 1} ^ {m + 1} k = \displaystyle \sum_{k = 1} ^ {m} k + m + 1 = \] \[\frac {m (m +1)} {2} + m + 1 = \] \[\frac {m ^ 2 + m + 2m + 2} {2} = \] \[\frac {m ^ 2 + 2m + m + 2} {2} = \] \[\frac {m (m + 2) + (m + 2)} {2} = \] \[\frac {(m + 1) (m + 2)} {2} \]

 \(\blacksquare \)

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