Solution:

Set $AA_1 \cap BB_1 = M$, $BB_1 \cap CC_1 = N$ and $CC_1 \cap AA_1 = P$. Using the standard notation for the angles of $\triangle ABC$, we have
$$\text{PMN = - B 1BC = - ( - ) =}$$
Analogously, we get $\text{MNP =}$ and $\text{NPM =}$, i.e. $\triangle NPM \sim \triangle ABC$.

Let $H$ be the orthocenter of $\triangle ABC$. The equalities $\text{HCC 1 = HBB 1 = HAA 1 = 90 -}$ imply that each of the quadrilaterals $ABMH$, $BCNH$ and $ACHP$ is cyclic. Therefore $\text{HMA = HBA = 90 -}$ and $\text{HPM = 180 - APH = ACH = 90 -}$, which implies that $H$ is the circumcenter of

$\triangle MNP$.

b)

We have proved that the points $A$, $B$, $M$ and $H$ are cyclic. Then the Sine theorem gives
$$\frac{MH}{\sin(90^\circ - \varphi)} = \frac{c}{\sin(180^\circ - \gamma)}$$
i.e. $MH = 2R \cos \varphi$. Since $MH$ is the circumradius of $\triangle MNP$, we conclude that
$$2 - \sqrt{3} = \frac{S_{MNP}}{S_{ABC}} = \frac{MH^2}{R^2} = 4 \cos^2 \varphi \Longleftrightarrow 2 \cos 2\varphi = -\sqrt{3}$$
Noting that $0 < 2\varphi < 180^\circ$ we get $2\varphi = 150^\circ$, i.e. $\varphi = 75^\circ$.