MathLabs

Problem 2

A non-isosceles triangle A1A2A3A_1A_2A_3 has sides a1a_1, a2a_2, a3a_3 with the side aia_i lying opposite to the vertex AiA_i. Let MiM_i be the midpoint of the side aia_i, and let TiT_i be the point where the inscribed circle of triangle A1A2A3A_1A_2A_3 touches the side aia_i. Denote by SiS_i the reflection of the point TiT_i in the interior angle bisector of the angle AiA_i. Prove that the lines M1S1M_1S_1, M2S2M_2S_2 and M3S3M_3S_3 are concurrent.
Step 3 of 8: Equal chords make T3T_3 an arc midpoint, forcing S1S2∥A1A2S_1S_2\parallel A_1A_2
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In plain words

A pure circle fact — tangent at an arc's midpoint is parallel to its chord — converts an algebraic length equality into the geometric parallelism we actually need.

S1T3=S2T3 ⇒ T3 bisects arc S1S2 ⇒ S1S2∥A1A2S_1T_3=S_2T_3 \ \Rightarrow\ T_3 \text{ bisects arc } S_1S_2 \ \Rightarrow\ S_1S_2 \parallel A_1A_2
Detailed analysis

Equal chords S1T3=S2T3S_1T_3=S_2T_3 on ω\omega mean T3T_3 bisects the arc S1S2S_1S_2. The tangent to ω\omega at T3T_3 is exactly the side A1A2A_1A_2 (by definition of T3T_3 as the touch point on that side), and the tangent at the midpoint of an arc is always parallel to the chord subtending the arc. Hence S1S2∥A1A2S_1S_2\parallel A_1A_2.