Library / /31 of 63

, 2019

Geometry Difficulty 8.2 Shortlist Prove it Turkey

Let $ABC$ be a triangle with $|AB| > |AC|$ . Let $D$ be the foot of the altitude drawn from $A$ to $BC$ , $K$ be the intersection of $AD$ and the internal bisector of the angle at $B$ , $M$ be the foot of the perpendicular drawn from $B$ to $CK$ and $N$ be the intersection of $BM$ and $AK$ . Let $T$ be the intersection of $AC$ with the line which passes through $N$ and is parallel to $DM$ .
Show that $BM$ is the internal bisector of the angle $\angle TBC$ .

Solution

Let $R$ and $S$ be the reflections of $C$ in the lines $BK$ and $BN$ , respectively. It is easy to see the following: $C, N, R$ are collinear, $A, B, R$ are collinear and $C, M, S$ are collinear.

By symmetry, $\angle BRK = \angle BSK = \angle KCB$ . But, $\angle KCB = \angle BNK$ , hence $B, K, N, R, S$ are concyclic. Let $X$ be an arbitrary point on the line which passes through $N$ and which is parallel to $DM$ . Then, $\angle NBK = \angle MDN = \angle XNK$ , hence $XN$ is tangent to the circle $BKNRS$ . By Pascal's theorem, the following three points are collinear: $A = NK \cap BR$ , $C = SK \cap NR$ and $XN \cap BS$ . Thus, the three lines $XN, BS, AC$ are concurrent. Hence, $BS$ passes through $T = XN \cap AC$ . Then, $\angle TBC$ is the same angle as $\angle SBC$ whose bisector is clearly $BM$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.