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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Algebraic proof of Pythagoras' theorem

Hello everyone. Today we will prove Pythagoras' theorem using elementary algebra. This proof was first appeared in China more than 2000 years ago. 

Theorem: Let aa and bb be the legs, and cc the hypotenuse of a right triangle. Then the equation c2=a2+b2c ^ 2 = a ^ 2 + b ^ 2 holds. 

Proof: 

Consider the following diagram consisting of a large square with side length a+ba + b and a small square with side length cc and notice four right triangles with legs a,ba, b and hypotenuse c c .

Let us denote the area of ​​the large square by PP . Then P=(a+b)(a+b)P = (a + b)(a + b) Let P1P_1 be the area of ​​one of the four right triangles, and P2P_2 the area of ​​a small square with side length cc . Then we have that: P1=ab2P_1 = \frac {ab} {2} P2=c2P_2 = c ^ 2 Since P=4P1+P2P = 4P_1 + P_2 we can write the following equation:
(a+b)(a+b)=4⋅ab2+c2(a + b) (a + b) = 4 \cdot \frac {ab} {2} + c ^ 2  From here we have that: a2+2ab+b2=2ab+c2a ^ 2 + 2ab + b ^ 2 = 2ab + c ^ 2  
That is, when we subtract 2ab2ab from both sides of the equation we get: c2=a2+b2c ^ 2 = a ^ 2 + b ^ 2  
■\blacksquare



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