My solution: Let $ D, E, F $ be the midpoint of $ BC, CA, AB $, respectively . Let $ K , G $ be the centroid of $ \triangle A_1B_1C_1, \triangle ABC $, respectively . Let $ A_2, B_2, C_2 $ be the midpoint of $ B_1C_1, C_1A_1, A_1B_1 $, respectively . Let $ D^* $ be the reflection of $ G $ in $ D $ . From the condition $ \Longrightarrow A_1A_2 \perp BC, B_1B_2 \perp CA, C_1C_2 \perp AB $ , so $ \triangle A_1B_1C_1 $ is the pedal triangle of the centroid $ K $ of $ \triangle A_1B_1C_1 $ WRT $ \triangle ABC $ $ \Longrightarrow K $ is the Symmedian point of $ \triangle ABC $ . Since $ \angle GD^*B=\angle D^*GC=\angle BAK+\angle KCB=\angle C_1B_1A_1 $ $ \angle D^*BG=\angle D^*BC+\angle CBG=\angle ACK+\angle KBA=\angle B_1A_1C_1 $ , so we get $ \triangle A_1B_1C_1 \sim \triangle BD^*G $ . Since $ BG=\tfrac{2}{3} BE, BD^*=CG=\tfrac{2}{3} CF, GD^*=AG=\tfrac{2}{3} AD $ , so $ \triangle A_1B_1C_1 $ is similar to the triangle whose side lengths are the medians of $ \triangle ABC $ . Q.E.D