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 of the sine theorem

Hello everyone. Today we will prove the sine theorem. We will use the method of direct proof as a proof method. This proof was first constructed by the Persian mathematician Tusi. 

Theorem: Let a,b,ca, b, c be the sides of any triangle, and RR the radius of the circumscribed circle around that triangle and let the angles α,β,γ\alpha, \beta, \gamma be the angles opposite the sides a,b,ca, b, c , respectively. Then the following equations hold: asin⁡α=bsin⁡β=csin⁡γ=2R\frac {a} {\sin \alpha} = \frac {b} {\sin \beta} = \frac {c} {\sin \gamma} = 2R  Proof: 

Now we will prove that the following equation holds: asin⁡α=2R\frac {a} {\sin \alpha} = 2R . The equations bsin⁡β=2R\frac {b} {\sin \beta} = 2R and csin⁡γ=2R\frac {c} {\sin \gamma} = 2R can be proved in an analogous way. 

Consider the following diagram showing the triangle △ABC\triangle ABC with the circumscribed circle of radius RR .

Since angles inscribed in a circle and subtended by the same chord are equal we have that ∠CA′B=α\angle CA'B = \alpha . We also know that the angle inscribed in a circle and subtended by the diameter of the circle is equal to 90∘90 ^ {\circ} , so it follows that ∠A′BC=90∘\angle A'BC = 90 ^ {\circ} . Further, according to the definition of the sine function, we have that a=2Rsin⁡∠CA′Ba = 2R \sin \angle CA'B , or a=2Rsin⁡αa = 2R \sin \alpha . Hence,  asin⁡α=2R\frac {a} {\sin \alpha} = 2R . 

■\blacksquare

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