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 cosine theorem

Hello everyone. Today we will prove the cosine theorem. We will use the method of direct proof as a proof method. This theorem was first formulated by the Persian mathematician Kashani. 

Theorem: Let \(a, b, c \) be the sides of any triangle and let the angles \(\alpha, \beta, \gamma \) be the angles opposite the sides \(a, b, c \), respectively. Then the following equations hold: \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \] \[b ^ 2 = a ^ 2 + c ^ 2-2ac \cos \beta \] \[a ^ 2 = b ^ 2 + c ^ 2-2bc \cos \alpha \] Proof: 

We will now prove that the first equation holds i.e. : \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \] Other two equations can be proved in an analogous way. 

There are three possible cases, and these are: \(\gamma \) is a right angle, \(\gamma \) is an acute angle, and \(\gamma \) is an obtuse angle. 

First case: \(\gamma = 90 ^ {\circ} \) 

According to the Pythagoras' theorem, we know that for a right triangle with hypotenuse \(c \) the following equation holds: \[c ^ 2 = a ^ 2 + b ^ 2 \] On the other hand, since \(\cos \gamma = \cos 90 ^ {\circ} = 0 \) it is true that: \[c ^ 2 = a ^ 2 + b ^ 2-0 \] \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos 90 ^ {\circ} \] \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \] Second case: \(\gamma <90 ^ {\circ} \) 

Consider the following diagram of an acute triangle \(\triangle ABC \) with height \(h \).

According to the Pythagoras' theorem, we can write the following two equations: \[a ^ 2 = h ^ 2 + p ^ 2 \] \[c ^ 2 = h ^ 2 + (bp) ^ 2 \] Combining these two equations we get: \[ c ^ 2 = a ^ 2-p ^ 2 + (bp) ^ 2 \] \[c ^ 2 = a ^ 2-p ^ 2 + b ^ 2-2bp + p ^ 2 \] \[c ^ 2 = a ^ 2 + b ^ 2-2bp \] Since \(p = a \cos \gamma \) it follows: \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \]Third case: \(\gamma> 90 ^ {\circ} \) 

Consider the following diagram of an obtuse triangle \(\triangle ABC \) with height \(h \).

According to the Pythagoras' theorem, we can write the following two equations: \[a ^ 2 = h ^ 2 + p ^ 2 \] \[c ^ 2 = h ^ 2 + (b + p) ^ 2 \] Combining these two equations we get: \[c ^ 2 = a ^ 2-p ^ 2 + (b + p) ^ 2 \] \[c ^ 2 = a ^ 2-p ^ 2 + b ^ 2 + 2bp + p ^ 2 \] \[ c ^ 2 = a ^ 2 + b ^ 2 + 2bp \] Since \(p = a \cos \left (180 ^{\circ}-\gamma \right) \) it follows: \[c ^ 2 = a ^ 2 + b ^ 2 + 2ab \cos \left (180 ^ {\circ}-\gamma \right) \] Applying identity \(\cos \left (180 ^ {\circ}-\gamma \right) = - \cos \gamma \) we get: \[c ^ 2 = a ^ 2 + b ^ 2-2ab \cos \gamma \]

\(\blacksquare\)

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