What is proof for pythagoras theorem?

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  • 9 years ago
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    Let ACB be a right-angled triangle with right angle CAB.

    On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.

    From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.

    Join CF and AD, to form the triangles BCF and BDA.

    Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.

    Showing the two congruent triangles of half the area of rectangle BDLK and square BAGF

    Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.

    Since AB is equal to FB and BD is equal to BC, triangle ABD must be congruent to triangle FBC.

    Since A-K-L is a straight line, parallel to BD, then parallelogram BDLK has twice the area of triangle ABD because they share the base BD and have the same altitude BK, i.e., a line normal to their common base, connecting the parallel lines BD and AL. (lemma 2)

    Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.

    Therefore rectangle BDLK must have the same area as square BAGF = AB^2.

    Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC^2.

    Adding these two results, AB^2 + AC^2 = BD × BK + KL × KC

    Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC

    Therefore AB^2 + AC^2 = BC^2, since CBDE is a square.

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  • Draw a square ABCD where each side length a + b units long. Draw point X on AB so that AX = a and BX = b. Draw point Y on BC so that BY = a and CY = b. Draw point Z on CD so that CZ = a and DZ = b. Draw point W on AD so that DW = a and AW = b. Now connect points XYZW.

    Since <A = <B = <C = <D = 90, note that by SAS congruence theorem that triangles:

    WAX, XBY, YCZ, ZDW are congruent to each other. Thus, corresponding sides and angles are congruent. Then XY = YZ = ZW = WX. Thus, all side lengths are equal. Let c be this common side lenght. Also, <AWX = <BXY = <CYZ = <DZW and <AXW = <BYX = <CZY = <DWZ.

    Note that <AXW + <AWX + <WAX = 180 since the sum of angles in a triangle is 180.

    Then <AXW + <AWX + 90 = 180 since <WAX is a right angle.

    Then <AXW + <AWX = 90.

    Note that <AXW + <WXY + <BXY = 180 since A,X,B are collinear.

    Then <AXW + <WXY + <AWX = 180 since <BXY = <AWX.

    Then <WXY + 90 = 180 since <AXW + <AWX = 90.

    Then <WXY = 90. Thus, <WXY is a right angle.

    Similarly, <XYZ = <YZW = <ZWX = 90. Thus, all angles are equal to 90. Now since all sides are equal and all angles are equal we see that XYZW is a square with side length c.

    Let's compare the areas.

    Area of square ABCD is (a + b)^2.

    Now look at it as a sum of the four triangles and square XYZW. Then the area is 4 * (1/2)(ab) + c^2.

    Then

    (a + b)^2 = 4*(1/2)(ab) + c^2

    a^2 + 2ab + b^2 = 2ab + c^2

    a^2 + b^2 = c^2

    DONE!

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  • 9 years ago

    Using 4 right triangles you arrange them in a specific way and then move them.

    Search youtube and you will probably find a video showing exactly how it is done.

    It is easier to see a visual than trying to explain it.

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  • 9 years ago

    that's very elaborate

    construct 3 squares on each side .

    drop normals on base from all corners

    deduce areas of those squares by manipulating areas of concerned triangles and trapezoids made with the base

    then it will turn out that area over hypoteneus square is sum of areas of squares on other two sides.

    that's the proof.

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  • 9 years ago

    Reach out to your geometry or maths sir/teacher. They'll teach you best

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  • 9 years ago

    a=5, b=,12

    a^2 + b^2 = c^2

    5^2 + 12^2 = c^2

    25 + 144 = c^2

    169 = c^2

    c^2 = 169

    c = √169

    c = 13

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  • 9 years ago
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    Lv 7
    9 years ago
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  • 9 years ago

    Visit http://www.mathsforums.org and get your answers.

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