Problem 2

A non-isosceles triangle

A_1A_2A_3

has sides

a_1

a_2

a_3

with the side

a_i

lying opposite to the vertex

A_i

. Let

M_i

be the midpoint of the side

a_i

, and let

T_i

be the point where the inscribed circle of triangle

A_1A_2A_3

touches the side

a_i

. Denote by

S_i

the reflection of the point

T_i

in the interior angle bisector of the angle

A_i

. Prove that the lines

M_1S_1

M_2S_2

and

M_3S_3

are concurrent.

Step 4 of 8: Cycling the argument: all three pairs of sides are parallel

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In plain words

The whole configuration has a 3-fold cyclic symmetry, so proving one pair of parallel sides and relabeling proves all three at once.

S_1S_2\parallel M_1M_2, \quad S_2S_3\parallel M_2M_3, \quad S_3S_1\parallel M_3M_1

Detailed analysis

By the Triangle Midline Theorem, $M_1M_2\parallel A_1A_2$ (the segment joining midpoints of two sides is parallel to the third side), so combined with Step 3, $S_1S_2\parallel M_1M_2$ . Repeating Steps 1–3 with the indices cycled $(1,2,3)\to(2,3,1)$ and $(3,1,2)$ gives $S_2S_3\parallel M_2M_3$ and $S_3S_1\parallel M_3M_1$ .