Prove that $\angle ABC <$ $ 84^\circ$. Angles A, B and C meet the circle at P, Q and R respectively.
I made the picture below, but what would the solution be?

 A: The figure shows $\triangle ABC$ with incenter $I$, inradius $r$, circumcenter $K$, and circum-half-radius $s:=|KS|$, where $S$ is the midpoint of $\overline{BC}$. (Since $\angle BKC=2\angle BAC=120^\circ$, we conclude $\triangle CKS$ is $30^\circ$-$60^\circ$-$90^\circ$, so that $|KS|$ is half the circumradius $|KC|$.) 

Now, the amount that $\angle B$ falls short of $90^\circ$ is the same the amount that $\angle C$ exceeds $30^\circ$; that's $\theta := \angle KCA$, and we have
$$\cos\theta = \frac{|AC|/2}{|KC|}=\frac{|AT|+|CT|}{4s}=\frac{|AT|+|CS|+|SR|}{4s}=\frac{(r+s)\sqrt{3}+p}{4s} \tag{1}$$
where
$$p^2 = r^2-(s-r)^2=s(2r-s) \tag{2}$$
From here, it's just a bit of algebra.
As in @dan_fulea's answer, we invoke Euler's formula, with circumradius $2s$ and incenter-to-circumcenter distance $r$:
$$r^2 = 4s(s-r) \quad\to\quad r = 2s (\sqrt{2}-1) \quad\to\quad p = s \sqrt{4 \sqrt{2}-5} \tag{3}$$
(where I've discarded negative $r$ and $p$ candidates) so that
$$\cos\theta = \frac14\left(\;(2\sqrt{2}-1)\sqrt{3}+\sqrt{4\sqrt{2}-5}\;\right) \tag{4}$$
It happens that $\theta = 6.09429\ldots^\circ$, from which the desired result follows. However, demonstrating this with a clean geometric argument eludes me. (See @dan's discussion.) Given that $\cos 6^\circ-\cos\theta \approx 0.00017$, there isn't a great deal of room to play.
A: $\begin{array}{} \text{Euler's relation} & IO=\sqrt{R(R-2r)} & IO=r=1 & R=1+\sqrt{2} \end{array}$
$\begin{array}{} sin(\frac{ π }{6})=\frac{1}{2} & AI=2 & AI^2=1+y_{A}^2 & A=(1,y_{A}=\sqrt{3}) \end{array}$
$\begin{array}{} {O}=λ ∩ ω & O=(h,k) \end{array}$
$\left\{ \begin{array}{} x^2+y^2=1 \\ (x-1)^2+(y-\sqrt{3})^2=(1+\sqrt{2})^2  \end{array}   \right.$
$\begin{array}{} h=\frac{1-\sqrt{2}+\sqrt{3+6\sqrt{2}}}{4} & k=\frac{-3+3\sqrt{2}+\sqrt{3+\sqrt{2}}}{4\sqrt{3}}  \end{array}$
$\begin{array}{} {C}=ψ∩t & \left\{ \begin{array}{} x=1 \\ (x-h)^2+(y-k)^2=(1+\sqrt{2})^2  \end{array}  \right. \end{array}$
$\begin{array}{} y_{C}=k+\sqrt{2h-h^2+2\sqrt{2}+2} & y_{C}=\frac{3\sqrt{3}+3\sqrt{6}+\sqrt{9+18\sqrt{2}}}{6}  \end{array}$
$\begin{array}{} AC=\sqrt{3}+y_{C} & AC=\frac{3\sqrt{3}+\sqrt{6}+\sqrt{1+2\sqrt{2}}}{2}  \end{array}$
$\begin{array}{} ϕ= \angle(AOB)& θ=\frac{ϕ}{2} & AO=OC=1+\sqrt{2} \end{array}$
$\begin{array}{} AC^2=AO^2+OC^2-2·AO·OC·cos(ϕ) & cos(ϕ)=1-\left( \frac{AC}{OC}  \right) ^2  \end{array}$
$cos(ϕ)=\frac{-7+4\sqrt{2}-5\sqrt{3+6\sqrt{2}}+3\sqrt{6+12\sqrt{2}}}{4}$
$\begin{array}{} θ=\angle(ABC) & θ=\frac{ϕ}{2}  \end{array}$
$θ=\frac{1}{2}arccos\left( \frac{-7}{4}+\sqrt{2}-\frac{5}{4}\sqrt{3+6\sqrt{2}}+\frac{3}{4}\sqrt{6-12\sqrt{2}}  \right)·\frac{180}{π}=83.905711$
$\begin{array}{} \text{checking} & θ<\frac{7π}{15} & ϕ=2θ<\frac{14π}{15} \end{array}$
$\begin{array}{} cos(ϕ)>cos(\frac{14π}{15}) & cos(ϕ)>cos(\frac{π}{15}) & cos(\frac{π}{15}>-cos(ϕ)) \end{array}$
$\frac{1}{8}\left( -1+\sqrt{5}+\sqrt{6(5+\sqrt{5}}  \right) >\frac{-1}{4}\left( -7+4\sqrt{2}-5\sqrt{3+6\sqrt{2}}+3\sqrt{6-12\sqrt{2}}  \right)$
how to prove it exactly?
$\begin{array}{} \text{True (numerical verification)} & \text{difference} & 6.896×10^{-4} \end{array}$

($λ$) Incircle of a $triangleABC$, (locus of point $O$) with center $I=(0,0)$ and radius $r=1$, $T=(1,0)$, ($t$) vertical line $x=1$ (locus of points $A$ and $C$), [$\frac{π}{3}$ angle construction] ($μ$) circle with center $I$ and radius=2, $A=(1,\sqrt{3})$ intersection $μ$ and $t$, $AI=2$, $D=(-2,0)$ intersection $μ$ and $x-axis$, $AD$, $\angle(IAT)=\frac{π}{6}$, $sin(\frac{π}{6})=\frac{1}{2}$, $\angle(DAI)=\frac{π}{6}$, ($ω$) circle with center $A$ and $R=1+\sqrt{2}$, (Euler's relation), $O=(h,k)$ intersection $λ$ and $ω$, ($ψ$) circle by $A$ with center in $O$ (locus points $A$, $B$ and $C$), $B$ intersection $AD$ and $ψ$, $C$ intersection $ψ$ and $t$, $BC$
