Dissection of Square into 8 Acute Triangles

Theorem

A square can be dissected into $8$ acute triangles.

Construction

Let $\Box ABCD$ be a square.

Let semicircles be constructed on $AD$ and $BC$ so as to meet in the center of $\Box ABCD$.

Let $E$ and $F$ be the midpoints of $CD$ and $AB$ respectively.

Let semicircles be constructed on $DE$ and $EC$

Let $G$ and $H$ be points on opposite sides of the straight line $EF$ such that both $G$ and $H$ are within the region bordered by the $4$ semicircles.

Then $\triangle AGF$, $\triangle FGH$, $\triangle HFB$, $\triangle BHC$, $\triangle CHE$, $\triangle EHG$, $\triangle DEG$, $\triangle DGA$ are $8$ acute triangles which compose $\Box ABCD$.

Proof

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