In acute \(\triangle ABC,\) \(D, E, F\) are the feet of perpendiculars from \(A, B, C\) to \(BC, CA, AB\) respectively. Let \(P, Q, R\) be the feet of perpendiculars from \(A, B, C\) to \(EF, FD, DE\) respectively. It turns out that lines \(AP, BQ, CR\) are concurrent at a point \(X\) within \(\triangle ABC.\) Then, \(X\) is the:
Details and assumptions
- This problem is not original; I got this from one of my friends. I don't know its original source.