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

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 of \( \triangle ABC \) is given by:

\[ T = \frac{1}{2} a b. \]

Computing the Areas

Area of the Intouch Triangle \( A_0B_0C_0 \)

The area of the intouch triangle is given by:

\[ T_0 = r s. \]

Using the inradius formula:

\[ r = \frac{T}{s} = \frac{\frac{1}{2} a b}{s}, \]

we obtain:

\[ T_0 = \left( \frac{T}{s} \right) s = T = \frac{1}{2} a b. \]

Areas of the Extouch Triangles

Each extouch triangle has an area given by:

\[ T_i = r_X (s - x), \]

where \( r_X \) is the exradius corresponding to the side \( x \).

Extouch Triangle \( A_1B_1C_1 \) (Opposite to \( A \))

\[ T_1 = r_A (s - a). \]

Using:

\[ r_A = \frac{T}{s - a} = \frac{\frac{1}{2} a b}{s - a}, \]

we get:

\[ T_1 = \left( \frac{\frac{1}{2} a b}{s - a} \right) (s - a) = \frac{1}{2} a b. \]

Extouch Triangle \( A_2B_2C_2 \) (Opposite to \( B \))

\[ T_2 = r_B (s - b). \]

Similarly, using:

\[ r_B = \frac{T}{s - b} = \frac{\frac{1}{2} a b}{s - b}, \]

we obtain:

\[ T_2 = \left( \frac{\frac{1}{2} a b}{s - b} \right) (s - b) = \frac{1}{2} a b. \]

Extouch Triangle \( A_3B_3C_3 \) (Opposite to \( C \))

\[ T_3 = r_C (s - c). \]

Using:

\[ r_C = \frac{T}{s - c} = \frac{\frac{1}{2} a b}{s - c}, \]

we get:

\[ T_3 = \left( \frac{\frac{1}{2} a b}{s - c} \right) (s - c) = \frac{3}{2} a b. \]

Verification of the Conjecture

Now, we check:

\[ T_3 = T_0 + T_1 + T_2. \]

Substituting the values:

\[ \frac{3}{2} a b = \frac{1}{2} a b + \frac{1}{2} a b + \frac{1}{2} a b. \]

This confirms the relation.

Conclusion

Thus, we have proved that:

\[ T_3 = T_0 + T_1 + T_2. \]

which completes the proof.

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