, 2006

In any obtuse triangle $XYZ$ with obtuse angle in $Z$ and ortocentre $W$, angles $XYZ$ and $XWZ$ are equal, as complementing the angle $YXW$ to $90$ degrees (fig. 1). Moreover, points $Y$ and $W$ lie in different half-planes determined by $XZ$.

Fig. 1

Let $P$, $Q$, $R$ be ortocentres of given triangles in that order. We will show $\not\leq PQR = 90^\circ$. Obviously all three triangles are obtuse in $C$. So $P$, $Q$, $R$ lie on extensions of altitudes going through $C$ to corresponding sides. Because of position of these sides it is also obvious the ray $CQ$ lies "between" rays $CP$ and $CR$, i.e. in angle $PCR$. So $\not\leq PQR = \not\leq RQC + \not\leq PQC$ (fig. 2). By the fact in the first paragraph $Q$, $R$ lie in

Fig. 2

the same half-plane determined by the line $BC$ and
$$\not\leq BEC = \not\leq BRC \quad \text{and} \quad \not\leq BDC = \not\leq BQC.$$
Angles $BEC$, $BDC$ are equal being inscribed angles with the same chord $BC$. Thus also $\not\leq BRC = \not\leq BQC = \omega$ and $BCRQ$ is cyclic. So $\not\leq RQC = \not\leq RBC = \varphi$. As $EC = r$ for inscribed angle we have $\not\leq EBC = 30^\circ$. Let $U$ be the foot on $BE$ in triangle $BEC$. Counting the angles in right triangle $BUR$ we get
$$\omega + \varphi + 30^\circ + 90^\circ = 180^\circ, \quad \text{i.e.} \quad \not\leq RQC = \varphi = 60^\circ - \omega = 60^\circ - \not\leq BDC.$$
In the same way we conclude $\not\leq PQC = 60^\circ - \not\leq DBC$. So we have (using the sum of angles in triangle $BCD$ is $180^\circ$)
$$\not\leq PQR = \not\leq RQC + \not\leq PQC = 120^\circ - (\not\leq BDC + \not\leq DBC) = \not\leq BCD - 60^\circ. \quad (1)$$
But also $BD = r$, thus $\not\leq BCD = 150^\circ$. Finally by (1) $\not\leq PQR = 90^\circ$.