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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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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 Area Relation in Right-Angled Triangle and Extouch Triangles

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Hello everyone. Today we will prove theorem about area relation in right-angled triangle and extouch triangles. Theorem Statement Let \( ABC \) be a right-angled triangle with the right angle at \( C \). Consider its associated special triangles: The intouch triangle \( A_0B_0C_0 \) formed by the contact points of the incircle. The extouch triangles \( A_1B_1C_1 \), \( A_2B_2C_2 \), and \( A_3B_3C_3 \), corresponding to the excircles touching \( BC \), \( AC \), and \( AB \) respectively. Denote their respective areas as \( T_0, T_1, T_2, \) and \( T_3 \). Then, the areas satisfy the relation: \[ T_3 = T_0 + T_1 + T_2. \] Proof Basic Notation Let: \( a, b, c \) be the sides of \( \triangle ABC \), where \( c \) is the hypotenuse. \( s \) be the semiperimeter: \[ s = \frac{a + b + c}{2}. \] \( r \) be the inradius (radius of the incircle). \( r_A, r_B, r_C \) be the exradii opposite to \( A, B, C \), respectively. The area ...
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Proof of the Theorem Regarding Golden Ratio

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 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 = \...
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