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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Algebraic proof of Pythagoras' theorem

Hello everyone. Today we will prove Pythagoras' theorem using elementary algebra. This proof was first appeared in China more than 2000 years ago. 

Theorem: Let \(a \) and \(b \) be the legs, and \(c \) the hypotenuse of a right triangle. Then the equation \(c ^ 2 = a ^ 2 + b ^ 2 \) holds. 

Proof: 

Consider the following diagram consisting of a large square with side length \(a + b \) and a small square with side length \(c \) and notice four right triangles with legs \(a, b \) and hypotenuse \( c \).

Let us denote the area of ​​the large square by \(P \). Then \[P = (a + b)(a + b) \] Let \(P_1 \) be the area of ​​one of the four right triangles, and \(P_2 \) the area of ​​a small square with side length \(c \). Then we have that: \[P_1 = \frac {ab} {2} \] \[P_2 = c ^ 2 \] Since \(P = 4P_1 + P_2 \) we can write the following equation:
\[(a + b) (a + b) = 4 \cdot \frac {ab} {2} + c ^ 2 \] From here we have that: \[a ^ 2 + 2ab + b ^ 2 = 2ab + c ^ 2 \] 
That is, when we subtract \(2ab \) from both sides of the equation we get: \[c ^ 2 = a ^ 2 + b ^ 2 \] 
\(\blacksquare \)



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