circleseuclidean-geometrygeometryplane-geometrytriangles
Let $FGH$ be a triangle with circumcircle $A$ and incircle $B$, the latter with touchpoint $J$ in side $GH$. Let $C$ be a circle tangent to sides $FG$ and $FH$ and to $A$, and let $D$ be the point where $C$ and $A$ touch, as shown here.
Prove that $\angle FGH = \angle GDJ$.
We begin by proving the claims made by Blue in the edit to the question.
Let $I$ be the incentre of $ABC$, and let $R$ be its circumradius. Since $O$ and $H$ lie inside $ABC$, the triangle must be acute.
The incircle is divided into three arcs by its points of contact with the sides of $ABC$. At least one of these arcs, say the one nearest vertex $A$, contains neither $O$ nor $H$. Thus when the rays $AO$ and $AH$ meet the incircle at $O$ and $H$, respectively, each of these rays is intersecting the incircle for the second time. Moreover, as in any triangle, $AI$ bisects angle $OAH$. It follows that the points $O$ and $H$ are symmetric about $AI$. In particular, $AH = AO = R$.
In any triangle, $\overrightarrow{AH} = 2\overrightarrow{OA'}$, where $A'$ is the midpoint of $BC$. Hence $OA' = R/2$. It follows from this that $\angle BOC = 120^{\circ}$, hence that $\angle BAC = 60^{\circ}$.
If we introduce $O'$ as in the figure (the reflection of $O$ through $A'$), then $O$, $B$ and $C$ belong to the circle with radius $R$ centred at $O'$. Since $\overrightarrow{AH} = \overrightarrow{OO'}$, the quadrilateral $OAHO'$ is a rhombus with side $R$. Consequently, $H$ also belongs to circle $BOC$.
If $J$ is the point halfway along arc $OH$ on circle $BOC$, then $BJ$ bisects $\angle OBH$, hence $J$ lies on $BI$. Similarly, $J$ lies on $CI$. Hence $J = I$, and $I$ lies on circle $BOC$. Since $AI$ bisects $\angle BAC$, it meets the circumcircle of $ABC$ again at $O'$, which is midway between $B$ and $C$.
Conversely, we carry out a construction corresponding to the above requirements. Start with a circle centred at $O$ with radius $1$. Mark two points $B$ and $C$ on the circle so that $\angle BOC = 120^{\circ}$. Let $O'$ be the reflection of $O$ through $BC$. Then $O'$ is on the circle. Now let $I$ be any point on circle $BOC$, on the same side of $BC$ as $O$. (We will specify $I$ further below.) Let $O'I$ cut $BO'C$ again at $A$. Let $H$ be the reflection of $O$ through $O'I$. Then reversing the arguments above, we find that $H$ is the orthocentre and $I$ the incentre of triangle $ABC$, and that $IH = IO$.
The only question that remains is how to choose $I$ on circle $BOC$ so that $IO$ is equal to the inradius of $ABC$. If we let $x$ be the inradius, then $x$ is the distance from $I$ to line $BC$. We also have $IO^2 = (1/2 - x)^2 + 1 - (x+1/2)^2 = 1-2x$. The condition $OI = x$ is equivalent to $x^2 = 1 - 2x$, or $x = \sqrt{2} - 1$.
Thus the construction can be completed by letting $I$ be a point of intersection of circle $BOC$ with a circle centred at $O$ with radius $\sqrt{2}-1$.
I'm not sure how to motivate this last step geometrically.
Summary of my construction Given two points $O$ and $O'$, write $R = OO'$. Construct the circles $K$ and $K'$ of radius $R$ centred at $O$ and $O'$, respectively. Let $B$ and $C$ be the points of intersection. Let $I$ be a point of intersection of $K'$ with the circle of radius $(\sqrt{2}-1)R$ centred at $O$. Then let $A$ be the point of intersection of $O'I$ with $K$.
Alternative construction (using $IA = 2IO$, proved by dxiv below) Instead of constructing $I$, construct $A$ directly by intersecting $K$ with the circle of radius $(2\sqrt{2} - 1)R$ centred at $O'$.
Summary of dxiv's construction Construct a triangle $AIO$ with $IO= r$, $IA = 2r$, $OA = (\sqrt{2}+1)r$. Let $K$ be the circle centred at $O$ passing through $A$. Construct angles of $30^{\circ}$ on either side of $AI$. Let $B$ and $C$ be the intersections with $K$ of the outer sides of these angles.
The area of the triangle ABC is
\begin{align} Area & = \frac12 r(AB + BC +CA) = \frac12 rAB \frac{\sin A+\sin B +\sin C}{\sin C} \\ &= \frac12 rAB \frac{4\cos \frac A2 \cos \frac B2\cos\frac C2 }{2\sin \frac C2\cos \frac C2} = rAB \frac{\cos \frac A2 \cos \frac B2 }{\cos \frac{A+B}2} = \frac{rAB }{1- \tan\frac A2\tan\frac B2 } \end{align}
Apply the Pythagorean theorem to the right triangle IFO, with $AB = 2R$
$$d^2 = (R-r)^2 - r^2 = R^2 - 2rR$$ $$\tan\frac A2\tan\frac B2= \frac r{R+d} \frac r{R-d}=\frac {r^2}{R^2-d^2} = \frac {r^2}{2rR} = \frac r{AB} $$ Thus $$Area = \frac{r AB }{1- \frac r{AB} } = \frac {rAB^2}{AB -r}$$
Best Answer
Notations:
Write $a:=GH$, $b:=HF$, $c:=FG$, and $s:=\frac{a+b+c}{2}$. Let $\Omega$ and $\omega$ be the circumcircle and the incircle of $FGH$, respectively. The circle internally tangent to $FG$, $FH$, and $\Omega$ is denoted by $\Gamma$. Suppose that $\Gamma$ intersects $HF$ and $FG$ at $P$ and $Q$, respectively. Denote by $\omega_a$ the excircle opposite to $F$ of $FGH$, which touches $GH$ at $T$. Extend $FT$ to meet $\Omega$ again at $S$. Finally, $\theta:=\angle GFD$.
Proof:
Let $i$ be the inversion at $F$ with radius $FP=FQ$. Then, $i(\Gamma)=\Gamma$, whereas $i(\Omega)$ is the tangent to $\Gamma$ at the point $E$, where $E$ is the second intersection between $\Gamma$ and $FD$. Suppose that $i(\Omega)$ meets $HF$ at $G'$ and $FG$ at $H'$. As $FG'H'$ and $FGH$ are similar triangles and $\Gamma$ is the excircle opposite to $F$ of $FG'H'$, it follows that $$\angle HFS=\angle HFT=\angle H'FE=\angle GFD=\theta\,.$$
Consequently, the minor arcs $HS$ and $GD$ of the circle $\Omega$ subtend the same angle $\theta$ at the circumference, so they are equal. Ergo, $HS=GD$. Since $TH=s-b=JG$ and $$\angle THS=\angle GHS=\angle GFS=\angle HFD=\angle HGD=\angle JGD\,,$$ we conclude that $GDJ$ and $HST$ are congruent triangles. Thence, $$\angle GDJ=\angle HST=\angle HSF=\angle FGH\,.$$
P.S.:
It can be shown, using Casey's Theorem, that the center of $\omega$ is the midpoint of $PQ$. Also, one can see that the internal angular bisector of $\angle FGH$ meets the line $DP$ at a point on $\Omega$, at which the tangent line $\ell_b$ is parallel to $HF$. Likewise, the internal angular bisector of $\angle GHF$ meets the line $DQ$ at a point on $\Omega$, at which the tangent line $\ell_c$ is parallel to $FG$. Finally, if $Z$ is the point of intersection between $\ell_b$ and $\ell_c$, then $Z$, $F$, $D$ are collinear.