Two non-intersecting circles, not lying inside each other, are drawn in the plane. Two lines pass through a point P which lies outside each circle. The first line intersects the first circle at A and A′ and the second circle at B and B′; here A and B are closer to P than A′ and B′, respectively, and P lies on segment AB. Analogously, the second line intersects the first circle at C and C′ and the second circle at D and D′. Prove that the points A, B, C, D are concyclic if and only if the points A′, B′, C′, D′ are concyclic.
Problem
Source: Juniors Problem 7
Tags: geometry unsolved, geometry
¬[ƒ(Gabriel)³²¹º]¼
30.07.2008 06:34
ABCD is cyclic iff $ \overline{PA} \cdot \overline{PB} = \overline{PC} \cdot \overline{PD}$. For the power of P wrt the first and the second circle we obtain $ \overline{PA} \cdot \overline{PA'} = \overline{PC} \cdot \overline{PC'}$ and $ \overline{PB} \cdot \overline{PB'} = \overline{PD} \cdot \overline{PD'}$; sobstituing these on $ \overline{PA} \cdot \overline{PC} = \overline{PB} \cdot \overline{PD}$ we obtain $ \overline{PA'} \cdot \overline{PB'} = \overline{PC'} \cdot \overline{PD'}$ that is true iff A'B'C'D' is cyclic.
parmenides51
20.12.2022 17:31
Two non-intersecting circles, not lying inside each other, are drawn in the plane. Two lines pass through a point $P$ which lies outside each circle. The first line intersects the first circle at $A$ and $A'$ and the second circle at $B$ and $B'$, here $A$ and $B$ are closer to $P$ than $A'$ and $B'$, respectively, and $P$ lies on segment $AB$. Analogously, the second line intersects the first circle at $C$ and $C'$ and the second circle at $D$ and $D'$. Prove that the points $A, B, C, D$ are concyclic if and only if the points $A', B', C', D'$ are concyclic.