This article demonstrates a generalized version of Thales’s theorem in ellipse.

Right triangle in circle

One can retrieve the original version of the Thales’ theorem either from:
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    In[]:=
    Thales' theorem
    »
    Thales' theorem
    GEOMETRIC SCENE
    Out[]=
    Thales' theorem
    In[]:=
    Thales' theorem
    GEOMETRIC SCENE
    ["Statement"]
    Out[]=
    If
    A
    ,
    B
    and
    C
    are points on a circle and the line
    AC
    passes through the center
    O
    of the circle, then the angle
    ∠ABC
    is a right angle.
    In[]:=
    Thales' theorem
    GEOMETRIC SCENE
    ["Scene"]//RandomInstance
    Out[]=
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    In[]:=
    Thales' theorem
    Input interpretation
    Thales’ theorem
    Definition
    An inscribed angle in a semicircle is a right angle.
    More information »
    Related terms
    inscribed angle
    |
    right angle
    |
    semicircle
    Subject classifications
    Show details
    MathWorld
    angles
    |
    circles
    MSC 2010
    51M04
    |
    51Nxx
    Associated person
    Thales

    Right triangle in an ellipse

    Given the black point sitting on the boundary of an ellipse with the length of major axis =1 and eccentricity =
    e
    . If the black point is also the 90-degree vertex of any inscribed right triangle, the hypotenuse of all such right triangles meet at one fixed point (orange). This fixed point is uniquely determined by the location of the black point. If the black point travels alone the ellipse, the trajectory of the orange point is also an ellipse. The inner ellipse comes from shrinking the outer one by a factor of
    2
    e
    2-
    2
    e
    .Notice this expression is less than one for
    0<e<1
    .
    In[]:=
    Plot[{1,ecc^2/(2-ecc^2)},{ecc,0,1},PlotLegends->"Expressions"]
    Out[]=
    1
    2
    ecc
    2-
    2
    ecc
    When
    e=0
    , the degenerated case is that the outer ellipse is circle. Then the fixed point and the inner ellipse becomes the origin, which is exactly the setup in Thales’ theorem: the mid point of hypotenuse of inscribed right triangle is the center of the reference circle. What’s more, if the black point travels counter-clockwise alone the outer ellipse, the orange point travels clockwise.
    The following code creates the animation in the post, where the variable
    k
    is slope of some hypotenuse:
    In[]:=
    DynamicModule{x0,y0,shrink},Manipulate​​shrink=ecc^2(2-ecc^2);​​x0=Cos[θ];​​y0=
    1-ecc^2
    Sin[θ];​​ParametricPlot​​Cos[t],
    1-ecc^2
    Sin[t],​​shrink*Cos[t],-shrink
    1-ecc^2
    Sin[t]​​,{t,-π,π},PlotRange->{{-1.05,1.05},{-1.05,1.05}},AspectRatio->1,ImageSize->600,​​Epilog->​​Table​​EdgeForm[{Thick,Hue[i/(10+n*5)+1/4]}],​​Transparent,Triangle{x0,y0},​​With{k=Tan[θ+i*Degree]},​​IfAbs[k]<1000.,(-(1-ecc^2)x0+k(kx0-2y0))(1-ecc^2+
    2
    k
    ),(-
    2
    k
    y0+(1-ecc^2)(-2kx0+y0))((1-ecc^2)+
    2
    k
    ),{x0,-y0},​​With{k=Tan[θ+i*Degree]},​​IfAbs[k]<1000.,(x0+(-1+
    2
    ecc
    )
    2
    k
    x0+2ky0)(1-(-1+
    2
    ecc
    )
    2
    k
    ),(y0+(-1+
    2
    ecc
    )k(2x0+ky0))(-1+(-1+
    2
    ecc
    )
    2
    k
    ),{-x0,y0}​​,{i,10,10+n*5,5},​​PointSize[0.02],Orange,Pointshrink*Cos[θ],-shrink
    1-ecc^2
    Sin[θ],​​PointSize[0.02],PointCos[θ],
    1-ecc^2
    Sin[θ]​​​​,​​{{ecc,0.3,"Eccentricity"},0.3,0.9},{{θ,-π,"Rt. Vertex"},-π,3π},{{n,0,"# of Rt."},0,8},TrackedSymbols:>{ecc,θ,n}

    CITE THIS NOTEBOOK

    Thales' theorem in an ellipse​
    by Shenghui Yang​
    Wolfram Community, STAFF PICKS, December 11, 2024
    ​https://community.wolfram.com/groups/-/m/t/3337073