Composition of Two Reflections Across Intersecting Lines

First, since $I$ lies on both $d_1$ and $d_2$, both reflections fix $I$. Take any point $M$ distinct from $I$, and let $M_1 = R_{d_1}(M)$ and $M' = R_{d_2}(M_1)$.

Because $d_1$ is the perpendicular bisector of $MM_1$ and $d_2$ is the perpendicular bisector of $M_1M'$, we have $IM = IM_1 = IM'$, so $M$ and $M' = R_{d_2}(M_1)$ lie on the same circle centered at $I$.

Let $\varphi_1$ be the directed angle from ray $\overrightarrow{IM}$ to $\overrightarrow{Id_1}$, and $\varphi_2$ be the directed angle from $\overrightarrow{Id_1}$ to $\overrightarrow{Id_2}$. Reflection across $d_1$ sends $\overrightarrow{IM}$ to $\overrightarrow{IM_1}$ by rotating through $2\varphi_1$, and reflection across $d_2$ sends $\overrightarrow{IM_1}$ to $\overrightarrow{IM'}$ by rotating through $2\varphi_2$. Adding the directed angles gives $(\overrightarrow{IM}, \overrightarrow{IM'}) = 2\varphi_1 + 2\varphi_2 = 2(\varphi_1 + \varphi_2) = 2\alpha$, which is independent of $M$. Hence every point is rotated around $I$ by $2\alpha$, proving the composition equals $Q_{(I, 2\alpha)}$.