Let $ABC$ be a triangle, let $P$ be the point $X(15)$ or $X(16)$. Let $A_0$ be a point on the plain such that:
$\angle A0BP=\angle A0CP = 60^0$; define $B_0,C_0$ cyclically. Show that $A_0B_0C_0$ is a equilateral triangle (and A0B0C0 lie on cirumcircle).
Let $A_1B_1C_1$ be a triangle such that $A_0B_0C_0$ are median triangle of $A_1B_1C_1$. Show that ABC and $A_1B_1C_1$ are perpective. Please see the figure attachment
Telv Cohl's proof:
Since $ \angle BCA_0=\angle BAP, \angle A_0BC=\angle PAC $ , so $ A_0 $ is the intersection of $ AP $ and $ \odot (ABC) $ . Similarly, $ B_0=BP \cap \odot (ABC), C_0=CP \cap \odot (ABC) $ .
Since $ \angle A_0B_0C_0=\angle PAB+\angle BCP=60^{\circ},\angle B_0C_0A_0=\angle PBC+\angle CAP=60^{\circ} $ , so we get $ \triangle A_0B_0C_0 $ is an equilateral triangle and $ \odot(A_0B_0C_0) $ is the incircle of $ \triangle A_1B_1C_1 $ , hence from Steinbart theorem we get $ AA_1, BB_1, CC_1 $ are concurrent .
$\angle A0BP=\angle A0CP = 60^0$; define $B_0,C_0$ cyclically. Show that $A_0B_0C_0$ is a equilateral triangle (and A0B0C0 lie on cirumcircle).
Let $A_1B_1C_1$ be a triangle such that $A_0B_0C_0$ are median triangle of $A_1B_1C_1$. Show that ABC and $A_1B_1C_1$ are perpective. Please see the figure attachment
Telv Cohl's proof:
Since $ \angle BCA_0=\angle BAP, \angle A_0BC=\angle PAC $ , so $ A_0 $ is the intersection of $ AP $ and $ \odot (ABC) $ . Similarly, $ B_0=BP \cap \odot (ABC), C_0=CP \cap \odot (ABC) $ .
Since $ \angle A_0B_0C_0=\angle PAB+\angle BCP=60^{\circ},\angle B_0C_0A_0=\angle PBC+\angle CAP=60^{\circ} $ , so we get $ \triangle A_0B_0C_0 $ is an equilateral triangle and $ \odot(A_0B_0C_0) $ is the incircle of $ \triangle A_1B_1C_1 $ , hence from Steinbart theorem we get $ AA_1, BB_1, CC_1 $ are concurrent .
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