D-Excircle of the Orthic triangle

Today we shall have a (very brief) look at the $D-$ excircle of the orthic triangle which appeared recently on the Vietnam Team Selection Test. This configuration can be transposed to the standard excircle configuration by shifting the reference triangle to the orthic triangle, but certain results are interpreted more naturally in an orthic setting.

As usual, a clickable GeoGebra link is attached to the following diagram.

A summary of the key results pertaining to this configuration,

An easy angle chase shows that is the incenter of and that , and are the , and excenters of the orthic triangle.
- Hence, the feet $X$ , $Y$ and $Z$ of the perpendiculars from $A-$ to the sides $EF$ , $DF$ and $DE$ respectively are the corresponding extouch points of the same triangle.
- Further, one can then observe that the lines $\overline{BC}$ and $\overline{YZ}$ are parallel.
The second intersection of line with and the second intersection of line with both lie on the excircle as well.
- It can then be seen that quadrilaterals $BXPE$ and $CXQF$ are both cyclic. Also, by Reim’s theorem note that the quadrilateral $BCPQ$ is cyclic.
- The circles $(BDY)$ and $(CDZ)$ can also be shown to intersect on the circumcircle of the orthic triangle $(DEF)$ .
The intersection of line with line and the intersection of with line can also be seen to lie on the circle via angle chase.
- Points $U$ and $V$ also lie on the circles $(QEZ)$ and $(PFY)$ respectively. Hence, Power of a Point at $X$ implies that point $X$ lies on the radical axis of circles $(PFY)$ and $(QEZ)$ , which is the statement of 2025 Vietnam TST Problem 3.
- Further, since quadrilaterals $BCEF$ and $BCUV$ are both cyclic, Reim’s theorem implies that lines $\overline{EF}$ and $\overline{UV}$ are parallel.
Quadrilaterals and are both cyclic, so it follows that also lies on the radical axis of circles and . This can also be shown quite simply using complex numbers, with as the unit circle.
- Hence, the second intersection points of circles $(PFY)$ and $(QEZ)$ with $YZ$ , $N$ and $L$ are symmetric with respect to point $M$ .
- Thus, point $M$ is the center of homothety mapping triangle $DYZ$ to triangle $HNL$ . Moreover, point $R$ is the center of homothety mapping the triangle $FL'S$ to $DYZ$ , and similarly.
- Some angle chasing can then be used to show that points $U$ , $V$ , $N$ , $L$ and $H$ all lie on a circle.
- It is also true the points $B$ , $C$ , $U$ , $V$ , $N$ and $L$ lie on the same conic, which gives us a plethora of collinearity/concurrency results via Pascal’s theorem.