# Six Concyclic Points via Antipedal Triangle

### What Might This Be About?

### Problem

$D$ a point in the plane of (non-degenerate) $\Delta ABC.$ Lines $d_a,$ $d_b,$ $d_c$ through $A,$ $B,$ and $C,$ are perpendicular to $AD,$ $BD,$ $CD,$ respectively. $A_1$ is the intersection of $d_b$ and $d_c;$ $B_1$ the intersection of $d_b$ and $d_c;$ $C_1$ that of $d_a$ and $d_b.$ Circles $(A_1),$ $(B_1),$ $(C_1),$ centered at $A_1,$ $B_1,$ $C_1$ and passing through $D,$ meet the side lines of $\Delta ABC$ in six points $A_b,$ $A_c,$ $B_a,$ $B_c,$ $C_a,$ $C_b,$ as shown below.

Prove that the six points are concyclic.

In passing, $\Delta A_{1}B_{1}C_{1}$ is known as the *antipedal triangle* of point $D$ with respect to $\Delta ABC.$

### Solution

Circles $(B_1)$ and $(C_1)$ intersect in $D$ and one other point, say $A'$ - its mirror image in the line of centers $B_{1}C_{1},$ or point $A,$ which is the same:

According to the Power of a Point Theorem, in circle $(B_1),$ $AB_a\times AB_c=AD\times AA' = AD^2.$

On the other hand, in circle $(C_1),$ $AC_a\times AC_b=AD\times AA' = AD^2,$ implying, by the the inverse of the Power of a Point Theorem, that points $B_a,$ $B_c,$ $C_a,$ $C_b$ are concyclic. Similarly, points $C_a,$ $C_b,$ $A_b,$ $A_c,$ are concyclic, and so are points $A_b,$ $A_c,$ $B_a,$ $B_c.$ Assuming the circles are distinct, the side lines of $\Delta ABC$ serve as radical axes of the pairs of the circles. This leads to a contradiction because the triangle is non-degenerate, whereas the three radical axes of three circles taken in pairs meet at their radical center. It follows that the three circles coincide, making all six points concyclic.

### Acknowledgment

The problem and its solution have been posted by Dao Thanh Oai at the CutTheKnotMath facebook page.

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