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Source of the problem
Start with hints
Draw a diagram. Construction: Complete the hexagon. Can you find some parallelograms in the picture? Particularly, can you prove \( BGCA_1 \) is a parallelogram?
(We will prove that the marked blue angles are equal) Notice that \( \angle ABC = A_1 C_1 B_1 \) (given) Also \( \angle A_1 C_1 B_1 = \angle A_1 B B_1 \) (subtended by the chord \(A_1 B_1\) at the circumference. Hence \( \angle ABC = \angle A_1 B B_1 \) Now substracting \( \angle CBB_1 \) from both side we have IMPORTANT: \( \angle A_1 B C = \angle ABB_1 \) Similarly notice that \( \angle BAC =\ C_1 B_1 A_1 \) (given) Also \( \angle C_1 B_1 A_1 = \angle C_1 A A_1 \) (subtended by the chord \(C_1 A_1 \) at the circumference. Hence \( \angle BAC = \angle C_1 A A_1 \) Now substracting \( \angle BAA_1 \) from both side we have IMPORTANT: \( \angle C_1 A B = \angle A_1AC\) Now \( \angle A_1 B C = \angle A_1 A C \) (subtended by the same segment \(A_1 C \) at the circumference.) And \( \angle C_1 A B = \angle C_1 C B \) ( subtended by the same segment \( C_1 B \) ) Thus \( \angle C_1 C B = \angle A_1 B C \) Thus alternate angles are equal making \( BA_1 \) parallel to GC. Thus in \( \Delta A_1MB \) and \( \Delta GMC \) we have BM = CM, \( \angle BMA_1 = \angle GMC \) and \( \angle C_1 C B = \angle A_1 B C \) Hence the triangles are congruent, making \( BA_1 = GC \)
Since one pair of opposite sides are equal and parallel hence the quadrilateral \(BGCA_1\) is a parallelogram.
Similarly G is the midpoint of \( BB_1 , CC_1 \) Can you conclude that G is the center?
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