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 neighboring Fibonacci numbers are coprime

Hello everyone. Today we will prove that the neighboring Fibonacci numbers are coprime. We will use the method of mathematical induction as a proof method. 

Theorem: Let \(F_n \) represent the nth Fibonacci number. Then: \[\forall n \ge 2, \quad \operatorname{NZD} \left (F_n, F_ {n + 1} \right) = 1 \] 

Proof: 

1. Base case (n = 2) 

\[ \operatorname{NZD} \left(F_2, F_{3} \right) = \operatorname{NZD} (1,2) = 1 \] 

2. Induction hypothesis (n = m) 

Suppose that: \[\operatorname { NZD} \left (F_m, F_ {m + 1} \right) = 1 \] 

3. Inductive step (n = m + 1) 

Using the assumption from the second step, we will prove that: \[\operatorname{NZD}\left(F_{m+1},F_{m+2}\right)=1\]Since the greatest common divisor of any natural numbers \(a \) and \(b \) is equal to the greatest common divisor of any linear combination of numbers \(a \) and \( b \) we have that \(\operatorname {NZD} (a, b) = \operatorname {NZD} (a, ba) \). With this in mind, we can write the following equation: \[\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left(F_ {m + 1} , F_ {m + 2} -F_ {m + 1} \right) \] Next,  by the definition of the Fibonacci sequence \(F_ {m + 2} = F_ {m} + F_ {m + 1} \) we have that: \[\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left (F_ {m + 1}, F_ {m} \right) \] \[\operatorname {NZD} \left (F_ {m + 1}, F_ {m + 2} \right) = \operatorname {NZD} \left (F_ {m}, F_ {m + 1} \right) = 1 \] 

\(\blacksquare \)

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