Proof Concerning Central Line X5X6X_5X_6 of Triangle

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Hello everyone. Today we will prove theorem about central line X5X6X_5X_6 of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let ABCDABCD be a golden rectangle where ABBC=ϕ \frac{AB}{BC} = \phi , and construct the square BCQP BCQP inside it. Reflect P P over D D to obtain E E . Then, the line EB EB coincides with the central line X5X6 X_5X_6 of triangle ABP ABP . Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: A=(0,0) A = (0,0) , B=(ϕx,0) B = (\phi x, 0) , C=(ϕx,x) C = (\phi x, x) , D=(0,x) D = (0, x) . The square BCPQ BCPQ ensures P=(ϕx,2x) P = (\phi x, 2x) . The reflection of P P across D D is E=(−ϕx,2x) E = (-\phi x, 2x) . 2. Compute the Nine-Point Center X5 X_5 The nine-point center X5 X_5 is the circumcenter of the medial triangle, which consists of the midpoints: M1=(0+ϕx2,0)=(ϕx2,0), 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 nn 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 nn the following formula holds: ∑k=1nk=n(n+1)2\displaystyle \sum_{k = 1} ^ nk = \frac {n (n + 1)} {2}  Proof: 

1. Base case (n = 1) ∑k=11k=1⋅(1+1)2\displaystyle \sum_{k = 1} ^ {1} k = \frac {1 \cdot (1 + 1)} {2} 1=1⋅(2)21 = \frac {1 \cdot (2)} {2} 1=221 = \frac {2} {2} 1=11 = 1  

2. Induction hypothesis (n = m) 

Suppose that: ∑k=1mk=m(m+1)2\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: ∑k=1m+1k=(m+1)(m+2)2\displaystyle \sum_{k = 1} ^ {m + 1} k = \frac {(m + 1) (m + 2)} {2}  So, ∑k=1m+1k=∑k=1mk+m+1=\displaystyle \sum_{k = 1} ^ {m + 1} k = \displaystyle \sum_{k = 1} ^ {m} k + m + 1 = m(m+1)2+m+1=\frac {m (m +1)} {2} + m + 1 = m2+m+2m+22=\frac {m ^ 2 + m + 2m + 2} {2} = m2+2m+m+22=\frac {m ^ 2 + 2m + m + 2} {2} = m(m+2)+(m+2)2=\frac {m (m + 2) + (m + 2)} {2} = (m+1)(m+2)2\frac {(m + 1) (m + 2)} {2}

 ■\blacksquare

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