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5 matchsticks, placed horizontally and vertically, divide the square into 2 equal parts as shown above. The 2 parts are equal in shape and size. Another obvious solution would be 1 vertical and 4 horizontal matchsticks based on the same principle.
But there is another clever solution! In what other way can you place the 5 matchsticks inside the square to divide it exactly into 2 parts equal in shape and size?
AC is exactly the length of 3 matchsticks and BC exactly 4 matchsticks. Applying the Pythagoras's theorem you can proof that AB is exactly the length of 5 matchsticks:
(AB)² = (BC)² + (AC)²
(AB)² = 16 + 9
AB = √25
AB = 5
(AB)² = (BC)² + (AC)²
(AB)² = 16 + 9
AB = √25
AB = 5
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