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 sine theorem

Hello everyone. Today we will prove the sine theorem. We will use the method of direct proof as a proof method. This proof was first constructed by the Persian mathematician Tusi. 

Theorem: Let \(a, b, c \) be the sides of any triangle, and \(R \) the radius of the circumscribed circle around that triangle and let the angles \(\alpha, \beta, \gamma \) be the angles opposite the sides \(a, b, c \), respectively. Then the following equations hold: \[\frac {a} {\sin \alpha} = \frac {b} {\sin \beta} = \frac {c} {\sin \gamma} = 2R \] Proof: 

Now we will prove that the following equation holds: \(\frac {a} {\sin \alpha} = 2R \). The equations \(\frac {b} {\sin \beta} = 2R \) and \(\frac {c} {\sin \gamma} = 2R \) can be proved in an analogous way. 

Consider the following diagram showing the triangle \(\triangle ABC \) with the circumscribed circle of radius \(R \).

Since angles inscribed in a circle and subtended by the same chord are equal we have that \(\angle CA'B = \alpha \). We also know that the angle inscribed in a circle and subtended by the diameter of the circle is equal to \(90 ^ {\circ} \), so it follows that \(\angle A'BC = 90 ^ {\circ} \). Further, according to the definition of the sine function, we have that \(a = 2R \sin \angle CA'B \), or \(a = 2R \sin \alpha \). Hence,  \(\frac {a} {\sin \alpha} = 2R \). 

\(\blacksquare \)

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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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