MATHEMATICS AND ORIGAMI

Jesús de la Peña Hernández

2. HAGA ́s THEOREM

Enunciation:

Let the square FIJH whose side measures one unit. If we fold vertex F over the mid-point of

JI, three right-angled triangles ∆(abc); ∆(xyz); ∆(def) are obtained in such a way that their

sides keep the proportion 3,4,5. Besides, beingx=^12 , it is also a=^23

3. HAGA ́s THEOREM EXTENSION

3.1 For any value of x, THE PERIMETER of ∆(abc) equals the sum of perimeters of

∆(xyz) and ∆(def).

Moreover, the perimeter of ∆(abc) is equal to half the perimeter of square FIJH.

3.2 KOHJI and MITSUE FUSHIMI

If x=^13 , it follows that a=^12 ; and if x=^14 , it ́s a=^25

4. HAGA ́s THEOREM DEMONSTRATION; likewise its extension will be demonstrated for
   any value of x.

Fig. 1 is a square whose side is equal to 1. F is folded over the upper side, being x the inde-
pendent variable. By so doing, fig. 2 is produced. In Fig. 3 BA is drawn perpendicular to JI through
B, so BA and EF are parallel. Fold BF is perpendicular to DE because DE is the perpendicular bisec-
tor of BF. Therefore ABEF is a quadrilateral having perpendicular diagonals and opposite vertices (B
and F) equidistant from the intersection point of those diagonals; moreover, opposite sides BA and
EF are parallel as said hereby. Consequently this quadrilateral is a rhomb.

JI

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a b

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y z

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a b

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1 23

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