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 Fibonacci numbers

Hello everyone. Today we will prove the formula for the sum of the first \(n \) Fibonacci numbers. We will use the method of mathematical induction as a proof method. 

Theorem: \(\forall n \in \mathbb {N} _0, \quad \displaystyle \sum_ {j = 0} ^ nF_j = F_ {n + 2} -1 \) 

Proof: 

1. Base case (n = 0)  \[\displaystyle \sum_ {j = 0} ^ 0F_j = F_ {2} -1 \] \[F_ {0} = F_ {2} -1 \] \[0 = 1-1 \] \[0 = 0 \] 

2. Induction hypothesis (n = m) 

Suppose that: \[\displaystyle \sum_ {j = 0} ^ mF_j = F_ {m + 2} -1 \] 

3. Inductive step (n = m + 1) 

Using the assumption from the second step, we will prove that: \[\displaystyle \sum_ {j = 0} ^ {m + 1} F_j = F_ {m + 3} -1 \] So,  \[\displaystyle \sum_ { j = 0} ^ {m + 1} F_j = \displaystyle \sum_ {j = 0} ^ {m} F_j + F_ {m + 1} = \] \[F_ {m + 2} -1 + F_ {m +1} = \] \[F_ {m + 1} + F_ {m + 2} -1 = \] \[F_ {m + 3} -1 \]

\(\blacksquare\)

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Proof Concerning Central Line \(X_5X_6\) of Triangle

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