Proof Concerning Central Line X5X6X_5X_6 of Triangle

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Hello everyone. Today we will prove theorem about central line X5X6X_5X_6 of triangle. Proof of the Central Line in a Golden Rectangle Construction Statement of the Theorem Let ABCDABCD be a golden rectangle where ABBC=ϕ \frac{AB}{BC} = \phi , and construct the square BCQP BCQP inside it. Reflect P P over D D to obtain E E . Then, the line EB EB coincides with the central line X5X6 X_5X_6 of triangle ABP ABP . Step-by-Step Proof 1. Define the Coordinates We assign coordinates as follows: A=(0,0) A = (0,0) , B=(ϕx,0) B = (\phi x, 0) , C=(ϕx,x) C = (\phi x, x) , D=(0,x) D = (0, x) . The square BCPQ BCPQ ensures P=(ϕx,2x) P = (\phi x, 2x) . The reflection of P P across D D is E=(−ϕx,2x) E = (-\phi x, 2x) . 2. Compute the Nine-Point Center X5 X_5 The nine-point center X5 X_5 is the circumcenter of the medial triangle, which consists of the midpoints: M1=(0+ϕx2,0)=(ϕx2,0), 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,RP, Q, R be three arbitrary points on the circumference of a circle k(O,r)k (O, r) . Then, ∠QOP=2∠QRP\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 : ∠POR+∠ORP+∠RPO=180∘\angle POR + \angle ORP + \angle RPO = 180 ^ {\circ} ∠QOP=180∘−∠POR\angle QOP = 180 ^{\circ} - \angle POR ∠ORP=∠QRP\angle ORP = \angle QRP Combining the first and second equation we get: ∠QOP=∠ORP+∠RPO\angle QOP = \angle ORP + \angle RPO Since triangle △POR\triangle POR is isosceles we have that ∠RPO=∠ORP\angle RPO = \angle ORP , so: ∠QOP=∠ORP+∠ORP\angle QOP = \angle ORP + \angle ORP ∠QOP=2∠ORP\angle QOP = 2 \angle ORP That is, since ∠ORP=∠QRP\angle ORP = \angle QRP we get: ∠QOP=2∠QRP \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,RS, O, R lie on the same line, based on the results from the first case, we conclude that ∠SOP=2∠SRP\angle SOP = 2 \angle SRP and ∠QOS=2∠QRS\angle QOS = 2 \angle QRS . Since ∠QOP=∠SOP+∠QOS\angle QOP = \angle SOP + \angle QOS we have that: ∠QOP=2∠SRP+2∠QRS\angle QOP = 2 \angle SRP + 2 \angle QRS ∠QOP=2(∠SRP+∠QRS)\angle QOP = 2 (\angle SRP + \angle QRS) ∠QOP=2∠QRP\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,RS, O, R lie on the same line, based on the results from the first case, we conclude that ∠POS=2∠PRS\angle POS = 2 \angle PRS and ∠QOS=2∠QRS\angle QOS = 2 \angle QRS . Since ∠QOP=∠QOS−∠POS\angle QOP = \angle QOS- \angle POS we have that: ∠QOP=2∠QRS−2∠PRS\angle QOP = 2 \angle QRS-2 \angle PRS ∠QOP=2(∠QRS−∠PRS)\angle QOP = 2 ( \angle QRS- \angle PRS) ∠QOP=2∠QRP\angle QOP = 2 \angle QRP  

■\blacksquare

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