
G4 Point $D$ lies on the side $[B C]$ of $\triangle A B C$. The circumcenters of $\triangle A D C$ and $\triangle B A D$ are $O_{1}$ and $O_{2}$, respectively and $O_{1} O_{2} \| A B$. The orthocenter of $\triangle A D C$ is $H$ and $A H=O_{1} O_{2}$. Find the angles of $\triangle A B C$ if $2 m(<C)=3 m(<B)$.


## Solution 1

![](https://cdn.mathpix.com/cropped/2024_06_05_b2c1a517537edbece730g-3.jpg?height=471&width=768&top_left_y=2069&top_left_x=655)

As $A D$ is the radical axis of the circumcircles of $\triangle A D C$ and $\triangle B A D$, we have that $O_{1} O_{2} \perp A D$, therefore $\widehat{D A B}=90^{\circ}$. Let $F$ be the midpoint of $[C D]$ and $[C E]$ be a diameter of the circumcircle of $\triangle A D C$. Then $E D \perp C D$ and $E A \perp C A$, so $E D \| A H$ and $E A \| D H$ since $A H \perp C D$ and $D H \perp A C$ (H is the orthocenter of $\triangle A D C$ ) and hence $E A H D$ is a parallelogram. Therefore $O_{1} O_{2}=A H=E D=2 O_{1} F$, so in $\triangle F O_{1} O_{2}$ with $\widehat{O_{1} F O_{2}}=90^{\circ}$ we have $O_{1} O_{2}=2 F O_{1} \Rightarrow \widehat{A B C}=\widehat{O_{1} O_{2} F}=30^{\circ}$.

Then we get $\widehat{A C B}=45^{\circ}$ and $\widehat{B A C}=105^{\circ}$.

