a_507_bc wrote: Let $ABC$ be a triangle with centroid $G$. Let $D, E, F$ be the circumcenters of triangles $BCG, CAG, ABG$. Let $X$ be the intersection of the perpendiculars from $E$ to $AB$ and from $F$ to $AC$. Prove that $DX$ bisects $EF$. Let $O$ be the circumcenter of $ABC$, let $M,N$ be the midpoints of $AB,EF$, let $P,Q$ be the second intersection of the circumcenters of $CAG,ABG$ respectivly with $BC$ and let $R,S$ be the feet of the altitudes from $E,F$ to $BC$. $D$ lies on the perpendicular bisector of $BC$. We have $MB\cdot MQ=MG\cdot MA=MC\cdot MP$. Thus $M$ is the midpoint of $PQ$ and the midpoint of $RS$ (since $R,S$ are the midpoints of $BQ,CP$). So $N$ lies on the perpendicular bisector of $BC$. We have $OF\perp AB\perp EX$ and $OE\perp AC\perp FX$. Thus $OEXF$ is a parallelogram or $O,E,X,F$ all lie on a line. The second option can be ruled out since $OF\perp AB,OE\perp AC$, so we would have $O=E$ or $O=F$ and $G$ would lie on the circumcircle of $ABC$ which contradicts that $G$ is inside $ABC$. So we conclude that $OEXF$ is a parallelogram and $N$ is the midpoint of $OX$ and $X$ lies on the perpendicular bisector of $BC$. So $DX$ is the perpendicular bisector of $BC$ and $N$ lies on $DX$.

Let $A',B',C'$ be the midpoints of $BC,AC,AB$ and let $O$ be the circumcenter of $\triangle ABC$. Clearly, $O = C'F \cap B'E$. Moreover, we have $EX \perp AB \perp FO$ and $FX \perp AC \perp EO$, which means that $EXFO$ is a parallelogram, as is $AB'A'C'$. We also have $FX \perp AC \parallel A'C'$, $EX \perp AB \parallel A'B'$, and $EF \perp AA'$, since $EF$ is the perpendicular bisector of $AG$. This means that the parallelograms $EXFO$, $AB'A'C'$ are similar and $OX \perp B'C' \parallel BC$. Hence $D,A',O,X$ all lie on a common line: the perpendicular bisector of $BC$. This line bisects $EF$, as desired.

Let the midpoint of $EF$ be $P$, and let the circumcenter of $ABC$ be $O$. Claim 1: $O, P, X$ are colinear. Proof: $E$ and $O$ both lies on the perpendicular bisector of $AC$, so $EO\perp AC$. By the problem's definition we have $FX\perp AC$, so $EO\parallel FX$. Similarly we have $FO\parallel EX$ so $OEXF$ is a parallelogram, and $OX$ will pass through the midpoint of $EF$ which is $P$. Claim 2: $DX\perp BC$ Proof: Let's prove that the triangles $ABC$ and $DFE$ are orthogonal to each other. Note that the perpendicular line from $B$ to $DE$ is the line through $B$ parallel to $CG$, and the perpendicular line from $C$ to $DF$ is the line through $C$ parallel to $BG$. The two lines intersects at the reflection of $G$ through midpoint of $BC$. The perpendicular line from $A$ to $EF$ is $AG$, which also pass through that point. So the two triangles are orthogonal, which mean the perpendicular line from $D$ to $BC$ pass through $X$. As $D, X, O, P$ are colinear we're done.

CrazyInMath wrote: Let the midpoint of $EF$ be $P$, and let the circumcenter of $ABC$ be $O$. Claim 1: $O, P, X$ are colinear. Proof: $E$ and $O$ both lies on the perpendicular bisector of $AC$, so $EO\perp AC$. By the problem's definition we have $FX\perp AC$, so $EO\parallel FX$. Similarly we have $FO\parallel EX$ so $OEXF$ is a parallelogram, and $OX$ will pass through the midpoint of $EF$ which is $P$. Claim 2: $DX\perp BC$ Proof: Let's prove that the triangles $ABC$ and $DFE$ are orthogonal to each other. Note that the perpendicular line from $B$ to $DE$ is the line through $B$ parallel to $CG$, and the perpendicular line from $C$ to $DF$ is the line through $C$ parallel to $BG$. The two lines intersects at the reflection of $G$ through midpoint of $BC$. The perpendicular line from $A$ to $EF$ is $AG$, which also pass through that point. So the two triangles are orthogonal, which mean the perpendicular line from $D$ to $BC$ pass through $X$. As $D, X, O, P$ are colinear we're done. bro can u pls explain what happens if two triangles are orthogonal or suggest some handout