Given a right nentagon A1A2 … A9, point O is its center. Prove that triangles A1OA4 and A1OA7 are equal

1) Let’s make a drawing
2) Since the hexagon A1A2 … A9 is regular, then ∠A1OA2 = ∠A2OA3 = … = ∠A8OA9 = ∠A9OA1 = 360o / 9 = 40o.
3) Then ∠A1OA4 = ∠A1OA2 + ∠A2OA3 + ∠A3OA4 = 40o * 3 = 120o.
4) ∠A1OA7 = ∠A1OA9 + ∠A9OA8 + ∠A8OA7 = 40o * 3 = 120o, that is, we get that ∠A1OA4 = ∠A1OA7.
5) Also, since the hexagon is regular, A1O = A7O = A4O.
6) Then we get that ∠A1OA4 = ∠A1OA7, A7O = A4O and A1O is a common side, therefore, triangles A1OA4 and A1OA7 are equal in criterion I, which was required to prove.