Why is the fundamental group of a triangle in a circle the free group on four generators?

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In lectures it was said that the fundamental group of the following shape, which we called $X_3$ (why?):

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is $\pi_1(X_3, P) \cong \mathbb{Z} * \mathbb{Z} * \mathbb{Z} * \mathbb{Z}$, i.e. the free group on four generators.

Why is this the case?

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Your space is homotopically equivalent to a wedge sum of $4$ circles. The result now follows from van Kampen.

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Path from Q to Q are a-a-a, a-b-a, a-b-b, b-b-b, any other combinations of a's and b's should be homotopy equivalent. With 4 non equivalent paths you get the free group on four generators