# Reflection in the Circumcenter

### What Might This Be About?

12 May 2014, Created with GeoGebra

### Problem

Points $A_1,$ $B_1,$ $C_1$ are projections of point $P$ on the sides $BC,$ $AC,$ $AB$ of $\Delta ABC .$ Points $A_2,$ $B_2,$ $C_2$ are the respective reflections of $A_1,$ $B_1,$ $C_1$ in the midpoints $M_a,$ $M_b,$ and $M_c$ of those sides.

Prove that the perpendiculars at $A_2,$ $B_2,$ $C_2$ to the sides $BC,$ $AC,$ $AB,$ respectively, are concurrent.

### Solution 1

Let $O$ be the circumcenter of $\Delta ABC.$ The projections (or pedal) points of $O$ are exactly the midpoints of the sides.

Define $Q$ to be the reflection of $P$ in $O.$ By Thales' Intercepts Theorem, the projections of $P$ and $Q$ on the sides are equidistant from the midpoints of the sides; $A_2,$ $B_2,$ $C_2$ are, therefore, the projections of $Q,$ and this proves the statement.

### Solution 2

The statement is a direct consequence of Carnot's theorem, according to which $AC_{1}^{2} - BC_{1}^{2} + BA_{1}^{2} - CA_{1}^{2} + CB_{1}^{2} - AB_{1}^{2} = 0.$ But, because $AC_{1}=BC_{2},$ $AC_{2}=BC_{1},$ etc., this is equivalent to

$AC_{2}^{2} - BC_{2}^{2} + BA_{2}^{2} - CA_{2}^{2} + CB_{2}^{2} - AB_{2}^{2} = 0,$

which implies the concurrency of the perpendiculars at the second set of the points.

### Acknowledgment

The problem has been posted by Dao Thanh Oai (Vietnam) at the CutTheKnotMath facebook page. He also supplied the second solution.

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