= Solution
Put $e_A=\varepsilon_{FA}F\eta_A$. This is a <natural transformation> $F\Rightarrow F$. Only the $G$-triangle $G\varepsilon_D\eta_{GD}=1_{GD}$ is assumed. Naturality gives two useful absorption identities:
$$
\begin{aligned}
Ge_A\eta_A
&=G\varepsilon_{FA}GF\eta_A\eta_A
=G\varepsilon_{FA}\eta_{GFA}\eta_A=\eta_A,\\
\varepsilon_D e_{GD}
&=\varepsilon_D\varepsilon_{FGD}F\eta_{GD}
=\varepsilon_DFG\varepsilon_DF\eta_{GD}=\varepsilon_D.
\end{aligned}
$$
For the second equality in the last line use naturality of $\varepsilon$ at $\varepsilon_D$, and then the assumed $G$-triangle. Naturality of $\varepsilon$ at $F\eta_A$ and of $\eta$ at $\eta_A$ now gives
$$
\begin{aligned}
e_A^2
&=\varepsilon_{FA}\varepsilon_{FGFA}FGF\eta_A F\eta_A\\
&=\varepsilon_{FA}\varepsilon_{FGFA}F\eta_{GFA}F\eta_A
=\varepsilon_{FA}e_{GFA}F\eta_A=e_A.
\end{aligned}
$$
Hence \b[$e$ is an idempotent in the functor category]: this is the <one-triangle adjunction idempotent>.
For the <splitting of an idempotent morphism>, suppose this <idempotent morphism> splits as <natural transformations> $r:F\Rightarrow H$ and $s:H\Rightarrow F$, with $sr=e$ and $rs=1_H$. Define
$$
\eta'_A=Gr_A\eta_A,\qquad \varepsilon'_D=\varepsilon_Ds_{GD}.
$$
The first absorption identity gives
$$
G\varepsilon'_D\eta'_{GD}
=G\varepsilon_DGe_{GD}\eta_{GD}=1_{GD}.
$$
For the other triangle, naturality of $s$ at $\eta'_A$ and of $\varepsilon$ at $r_A$ gives
$$
\begin{aligned}
\varepsilon'_{HA}H\eta'_A
&=\varepsilon_{HA}s_{GHA}H\eta'_A
=\varepsilon_{HA}F\eta'_A s_A\\
&=\varepsilon_{HA}FGr_A F\eta_A s_A
=r_A\varepsilon_{FA}F\eta_A s_A
=r_Ae_As_A=1_{HA}.
\end{aligned}
$$
Thus the <triangle identities for an adjunction> prove $H\dashv G$.
Conversely, suppose $H\dashv G$, with unit $\psi:1_{\mathcal C}\Rightarrow GH$ and counit $\varphi:HG\Rightarrow1_{\mathcal D}$. The unit has target $GH$, as its type requires. Define
$$
\boxed{s_A=\varphi_{FA}H\eta_A:HA\to FA,\qquad
r_A=\varepsilon_{HA}F\psi_A:FA\to HA.}
$$
These are <natural transformations>. Transposition under $H\dashv G$ gives $Gs_A\psi_A=\eta_A$. Independently, naturality of $\eta$ at $\psi_A$ and the assumed $G$-triangle give
$$
Gr_A\eta_A=G\varepsilon_{HA}GF\psi_A\eta_A
=G\varepsilon_{HA}\eta_{GHA}\psi_A=\psi_A.
$$
The transpose of $r_As_A$ is therefore $Gr_AGs_A\psi_A=\psi_A$, the transpose of $1_{HA}$. Injectivity of the hom-set <bijection> implies $r_As_A=1_{HA}$. Naturality of $\varepsilon$ at $s_A$ gives
$$
s_Ar_A=\varepsilon_{FA}FGs_AF\psi_A
=\varepsilon_{FA}F(Gs_A\psi_A)=\varepsilon_{FA}F\eta_A=e_A.
$$
Consequently \b[$G$ has a left adjoint if and only if $e$ splits]. This is the criterion for <splitting a one-triangle adjunction idempotent>. The argument gives both the explicit splitting and the new unit and counit, without assuming the other triangle for $F$.
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