Proof Concerning Central Line \(X_5X_6\) of Triangle

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:

\[ M_1 = \left(\frac{0 + \phi x}{2}, 0\right) = \left(\frac{\phi x}{2}, 0\right), \] \[ M_2 = \left(\frac{\phi x + \phi x}{2}, \frac{0 + 2x}{2} \right) = \left(\phi x, x\right), \] \[ M_3 = \left(\frac{0 + \phi x}{2}, \frac{0 + 2x}{2} \right) = \left(\frac{\phi x}{2}, x\right). \]

The circumcenter (nine-point center) is found by averaging the x-coordinates and y-coordinates:

\[ X_5 = \left(\frac{M_1 + M_2 + M_3}{3}\right) = \left(\frac{3\phi x}{4}, \frac{x}{2}\right). \]

3. Compute the Symmedian Point \( X_6 \)

The symmedian point \( X_6 \) is known from standard formulas for right triangles:

\[ X_6 = (\phi x, x). \]

4. Find the Equation of Line \( EB \)

The slope of \( EB \) is calculated as:

\[ m_{EB} = \frac{0 - 2x}{\phi x - (-\phi x)} = \frac{-2x}{2\phi x} = \frac{-1}{\phi}. \]

Using the point-slope form with \( B(\phi x, 0) \):

\[ y - 0 = \frac{-1}{\phi} (x - \phi x), \] \[ y = -\frac{1}{\phi} x + x. \]

5. Find the Equation of Line \( X_5 X_6 \)

The slope of \( X_5 X_6 \) is calculated using:

\[ m_{X_5X_6} = \frac{x - \frac{x}{2}}{\phi x - \frac{3\phi x}{4}} = \frac{\frac{x}{2}}{\frac{\phi x}{4}} = \frac{1}{\phi}. \]

6. Compare the Two Lines

Since:

\[ m_{EB} = -\frac{1}{\phi}, \quad m_{X_5X_6} = \frac{1}{\phi}. \]

The slopes are opposite in sign, meaning that the two lines are reflections across the x-axis.

7. Geometric Confirmation

The final confirmation comes from **reflection symmetry**:

  • \( X_5X_6 \) passes through the **reflection of \( P \)**.
  • \( E \) is also the **reflection of \( P \)**.
  • Thus, **\( EB \) and \( X_5X_6 \) must coincide**.

8. Conclusion

Therefore, we conclude:

\[ EB = X_5 X_6. \]

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