Dao's blog: 22-X(15) or X(16) and three antipode lie on a circle

Let $ABC$ be a triangle. Construct three equilateral triangle on the side: $\triangle ABC_1, \triangle BCA_1, \triangle CAB_1$ ( either all outward, or all inward).

Construct circle through $A$ , and through midpoints of $AC_1$ , $AB_1$ , center of circle is $C_A$
Construct circle through $B$ , and through midpoints of $BC_1$ , $BA_1$ , center of circle is $C_B$
Construct circle through $C$ , and through midpoints of $AC_1$ , $AB_1$ , center of circle is $C_C$ .

Prove that:
- $A'$ is antipode of $A$ with $(C_A)$
$B'$ is antipode of $B$ with $(C_B)$
$C'$ is antipode of $C$ with $(C_C)$
Four points: $X_{16}$ or $X_{15}$ and A',B',C' lie on a circle

Dao's blog

Đào Thanh Oai

Thứ Sáu, 12 tháng 9, 2014

22-X(15) or X(16) and three antipode lie on a circle

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