Monday, December 22, 2008

Bottema's Theorem: Triangle and Squares

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Draw squares ABDE and BCFG on sides AB and BC of a triangle ABC. Then the midpoint M of EF is independent of B and the triangle AMC is an isosceles right triangle.


Bottema's Theorem: Triangle and Squares.
Continue reading at:
gogeometry.com/geometry/bottema_theorem_triangle_square.htm

Posted by Antonio Gutierrez at 9:32 AM
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4 comments:

  1. natasJanuary 26, 2009 at 9:25 AM

    this's actually not hard to prove.
    T: midpoint of AC
    Let M,N,P be the points so that EM,FN, BP is perpendicular to AC.
    easily have: EM+FN=AC=2MT => MT // EM=> M is independant

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  2. APOSTOLIS MANOLOUDISAugust 25, 2017 at 1:22 PM

    Problem 1344
    Let EE’,BB’,MM’,FF’ perpendicular at AC then 1.tri ABB’=tri EE’A=>EE’=AB’,EA=BB’,
    2.tri BB’C=tri CFF’=>BB’=CF’,B’C=FF’.At the trapezoide EE’F’F MM’ is median so
    MM’=(EE’+FF’)/2=(AB’+B’C)/2=AC/2 .Is M’ medpoint AC and E’F’ so tri AMC is
    an isosceles right.In the extention of the B’B to the B point K such that BK=AC.
    Then tri BKG=tri CAB(S,A,S)(DK=BC=BG=>DKGB=parallelogram.If the DG intersect the KB at H,
    Then BH=BK/2=AC/2=MM’,(BH//MM’)=>BHMM’= parallelogram =>BM’=HM.
    Now tri BHG=tri BCM=>BM’=HG=HD=HM=> <LBD+<KDB=90.
    So the MH is perpendicular at DG.Therefore tri DMG is isosceles and right.
    APOSTOLIS MANOLOUDIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE


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  3. Benjamin LeisAugust 25, 2017 at 4:02 PM

    Square off both squares so they are each surrounded by 4 congruent triangles.

    Let x be the height of E, y be the height of B and z be the height of F. Then one set of triangles has lengths x,y, AB and the other ones z,y,BC.

    Set the lower left corner to (0,0) for convenience.

    From there M is at coordinates: y + (x+z)/2, (x+z)/2. Its clear the height is dependent only on x+z i.e. AC likewise the length is as well. As you move C only y can change and its counterbalanced on both sides.

    This also means if we draw a perpendicular down from M its height is x+z/2 and its located at x+z/2 i.e. the midpoint between AC. So that forms 2 isosceles right triangles that combine to form a larger one ACM.

    You can compute the coordinates of D and G similarly and show they are inverted slopes away from M: deltaX = y + (z - x)/2 delta y = +-(z - x)/2 - y So DGM is
    also a right isosceles.

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  4. Peter TranAugust 25, 2017 at 9:21 PM

    https://goo.gl/photos/5ZVEeXc42HDNTC6b7

    Let P, N, K and Q are the projection of points E, M, B and F over AC ( see sketch)
    Let DC meet AG at R
    1,2 Since M is the midpoint of EF so N is the midpoint of PQ
    Observe that triangle BCK and CFQ are congruent ..( case ASA) so CQ=BK and FQ=CK
    Simillarly BAK are congruent to AEP ( case ASA) so AP=BK=CQ and EP=AN
    N is the midpoint of PQ and AP=BK=CQ => N is the midpoint of AC
    In trapezoid EFQP we have MN= ½(EP+FQ)= ½(AK+KC)= 1/2AC
    So triangle AMC is an isoceles right triangle and position of M is independent to position of B
    3. Note that triangle DBC is the image of ABG in the rotational transformation center B , rotational angle= 90 so DC=AG and DC ⊥AG
    Quadrilateral AMRC is cyclic => ∠ (MAR)= ∠ (MCR)
    Triangle AMG are congruent to CMD ..( case SAS)
    And triangle CMD is the image of AMG in the rotational transformation center M , rotational angle= 90 so MD=MG and MD⊥MG
    So triangle DMG is an isoceles right triangle

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