IRL_ABooklet_2023

Let $E$ be the midpoint of $CA$ and $F$ the intersection point of $AB$ and the bisector of $\angle ACB$. We will use (the converse of) Ceva's Theorem to prove that $AD$, $BE$, $CF$ are concurrent.



We first note that $A$, $B$, $D$, $H$ are concyclic because of the right angles at $D$ and $H$. The Intersecting Secants Theorem then implies $$\begin{array}{rl}|CD|\cdot |CB|& =|CH|\cdot |CA|,\text{ i.e.}\\ \frac{|CH|}{|DC|}& =\frac{|CB|}{|CA|}\end{array}$$ Using $|CH|=|BD|$ and that the angle bisector $CF$ divides $AB$ in the ratio of the adjacent sides, we obtain $$\frac{|BD|}{|DC|}=\frac{|FB|}{|AF|},\text{ or }\frac{|BD|}{|DC|}\cdot \frac{|AF|}{|FB|}=1.$$ Because $|CE|=|EA|$, we finally obtain $$\frac{|BD|}{|DC|}\cdot \frac{|CE|}{|EA|}\cdot \frac{|AF|}{|FB|}=1$$ which implies that $AD$, $BE$, $CF$ are concurrent.