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 that the central angle of the circle is equal to twice the corresponding inscribed angle

Hello everyone. Today we will prove that the central angle of a circle is equal to twice the corresponding inscribed angle of the circle. We will use the method of direct proof as a proof method. 

Theorem: The central angle of a circle is equal to twice the corresponding inscribed angle of the circle. 

Proof: 

We can reformulate the statement as follows: 

Let \(P, Q, R \) be three arbitrary points on the circumference of a circle \(k (O, r) \). Then, \(\angle QOP = 2 \angle QRP \). 

We will construct the proof by proving three separate possible cases: 

First case: The center of the circle is on a side of the inscribed angle. 

We can write the following equations : \[\angle POR + \angle ORP + \angle RPO = 180 ^ {\circ} \] \[\angle QOP = 180 ^{\circ} - \angle POR \] \[\angle ORP = \angle QRP \] Combining the first and second equation we get: \[\angle QOP = \angle ORP + \angle RPO \] Since triangle \(\triangle POR \) is isosceles we have that \(\angle RPO = \angle ORP \), so: \[\angle QOP = \angle ORP + \angle ORP \] \[\angle QOP = 2 \angle ORP \] That is, since \(\angle ORP = \angle QRP \) we get: \[ \angle QOP = 2 \angle QRP \] 

Second case: The center of the circle is in the inner area of ​​the inscribed angle. 

Consider the following diagram:


Since the points \(S, O, R \) lie on the same line, based on the results from the first case, we conclude that \(\angle SOP = 2 \angle SRP \) and \(\angle QOS = 2 \angle QRS \) . Since \(\angle QOP = \angle SOP + \angle QOS \) we have that: \[\angle QOP = 2 \angle SRP + 2 \angle QRS \] \[\angle QOP = 2 (\angle SRP + \angle QRS) \] \[\angle QOP = 2 \angle QRP \] 

Third case: The center of the circle is outside the inner area of ​​the inscribed angle. 

Consider the following diagram:


Since the points \(S, O, R \) lie on the same line, based on the results from the first case, we conclude that \(\angle POS = 2 \angle PRS \) and \(\angle QOS = 2 \angle QRS \) . Since \(\angle QOP = \angle QOS- \angle POS \) we have that: \[\angle QOP = 2 \angle QRS-2 \angle PRS \] \[\angle QOP = 2 ( \angle QRS- \angle PRS) \] \[\angle QOP = 2 \angle QRP \] 

\(\blacksquare\)

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