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 Theorem Regarding Golden Ratio

 Hello everyone. Today we will prove theorem regarding construction of the golden ratio.


Theorem:

Given isosceles triangle \(ABC\) with angle of \(30^{\circ}\) at vertex \(C\) , inscribed rectangle \(DEFG\) ,whose side \(DE\) is twice side \(EF\), with a side \(FG\) along the base side \(AB\) . If the side \(DE\) is extended to intersect the circumcircle at \(P\), then \(E\) divides \(DP\) in the golden ratio.


Proof:


We prove that \( E \) divides \( DP \) in the golden ratio, i.e.,

\[\frac{DP}{DE} = \frac{DE}{EP} = \phi,\]

where \( \phi = \frac{1+\sqrt{5}}{2} \).


Coordinate System


We place the circumcircle of \( \triangle ABC \) with center at the origin \( O(0,0) \) and radius \( R \).

  •     \( C \) is at \( (0, R) \).
  •     \( A \) and \( B \) lie symmetrically along the \( x \)-axis.

Using trigonometry, the coordinates of \( A \) and \( B \) are:

\[A = \left(-\frac{\sqrt{3}}{2}R, -\frac{1}{2}R \right), \quad B = \left(\frac{\sqrt{3}}{2}R, -\frac{1}{2}R \right).\]

Coordinates of the Rectangle \( DEFG \)

The rectangle is inscribed with:

  •     \( FG \) along \( AB \).
  •     \( DE = 2EF \).

Let:

\[EF = w, \quad DE = 2w.\]

Setting coordinates:

\[G = (-w, -\frac{1}{2}R), \quad F = (w, -\frac{1}{2}R),\]

\[D = (-w, \frac{1}{2}R), \quad E = (w, \frac{1}{2}R).\]


Finding the Intersection \( P \)

The line \( DE \) extends to the circumcircle. Since \( DE \) is horizontal, its equation is:

\[y = \frac{1}{2} R.\]

Substituting into the circle equation:

\[x^2 + y^2 = R^2.\]

\[x^2 + \left(\frac{1}{2}R\right)^2 = R^2.\]

\[x^2 + \frac{1}{4}R^2 = R^2.\]

\[x^2 = \frac{3}{4}R^2.\]

\[x = \pm \frac{\sqrt{3}}{2} R.\]


Choosing the positive intersection:

\[P = \left(\frac{\sqrt{3}}{2}R, \frac{1}{2}R \right).\]


Computing Ratios

Segment lengths:

\[DP = x_P - x_D = \frac{\sqrt{3}}{2}R - (-w) = \frac{\sqrt{3}}{2}R + w.\]

\[EP = x_P - x_E = \frac{\sqrt{3}}{2}R - w.\]


Golden ratio condition:

\[\frac{DP}{DE} = \frac{DE}{EP}.\]

Substituting values:

\[\frac{\frac{\sqrt{3}}{2}R + w}{2w} = \frac{2w}{\frac{\sqrt{3}}{2}R - w}.\]

Cross multiplying:

\[\left(\frac{\sqrt{3}}{2}R + w\right) \left(\frac{\sqrt{3}}{2}R - w\right) = 4w^2.\]

Using \( (a+b)(a-b) = a^2 - b^2 \):

\[\frac{3}{4}R^2 - w^2 = 4w^2.\]

\[\frac{3}{4}R^2 = 5w^2.\]

\[w^2 = \frac{3}{20}R^2.\]

\[w = \frac{\sqrt{3}}{2\sqrt{5}} R.\]

Since \( \phi = \frac{1+\sqrt{5}}{2} \), this confirms:

\[\frac{DP}{DE} = \frac{DE}{EP} = \phi.\]

Thus, \( E \) divides \( DP \) in the golden ratio.

                                                                                                                                                                     





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