Library / /40 of 96

, 2021

Geometry Difficulty 8.2 Shortlist Prove it Baltic Way

Let $AD$ be the $A$ -altitude of an acute-angled triangle $ABC$ . The internal bisector of angle $DAC$ intersects $BC$ at $K$ . Let $L$ be the projection of $K$ onto $AC$ . Let $M$ be the intersection point of $BL$ and $AD$ . Let $P$ be the intersection point of $MC$ and $DL$ . Prove that $PK \perp AB$ .

Solution

Solution. Since $\angle IFK = 90^\circ$ , then $IK$ is the diameter of the circumcircle of $CFI$ , hence also $\angle ICK = 90^\circ$ . Similarly is $IL$ the diameter of the circumcircle of $BGI$ and $\angle IBL = 90^\circ$ . Therefore are the lines $CK$ and $GL$ parallel, also $BL$ and $FK$ are parallel.
Let the lines $CK$ and $BL$ intersect at $D$ , as seen in figure 17. From the above we get that $DKAL$ is a parallelogram. Note that $D$ is the excenter with respect to the vertex $A$ of the triangle $ABC$ , since the lines $BL$ and $CK$ are perpendicular to the corresponding internal angle bisectors. The excenter lies on the internal angle bisector $AI$ , hence $AI$ bisects the diagonal $KL$ .

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 reproduced verbatim; metadata (topic, difficulty) added by this project.