Why is this angle always less than $49°$?

A complete geometric proof

Assume, WLOG, that $\overline{BG} = 2$. Starting from this segment, produce it to $M$, so that $GM \cong \frac12 BG$. Clearly $M$ is the midpoint of $AC$. Since we want a fixed angle $\measuredangle MCG$, we must take $C$ on a circlular arc $\gamma$ with center on the axis of segment $GM$.

Angle $\angle GBC$ is maximized if we choose $C$ so that $BC$ is tangent to the circlular arc $\gamma$ (see figure below, where, additionaly, a non optimal solution $A'BC'$ is shown).

Since $\angle BCG \cong \angle BMC$ (Inscribed Angle Theorem), by AAA criterion we have $BCG \sim BCM$. Thus, recalling that $\overline{BG} = 2$ and $\overline{BM} = 3$, we get $$\frac{\overline{BM}}{\overline{BC}}=\frac{\overline{BC}}{\overline{BG}},$$ i.e. $$\overline{BC}^2 = 6.$$ The same similarity now gives us $$\frac{\overline{CM}}{\overline{CG}} = \frac3{\sqrt 6}.$$ Since $AC \cong 2 CM$ we immediatly get $$\boxed{\overline{AC} = \sqrt 6\ \overline{CG}},\tag{1}\label{1}$$ as per Blue's answer.

---

For the sake of completeness: without using \eqref{1} explicitely, we can take advantage directly of tangency condition discussed above, and express $x=\measuredangle GBC$ in terms of $\theta = \measuredangle GCA$ as follows.

First note that $$2 \measuredangle GCB + \theta +x = 180^\circ,$$ using the above similarity, and the interior angles of $BMC$. Hence $$\measuredangle GCB = 90^\circ-\frac{\theta +x}2.$$ Now we get an equation by expressing $\overline{GC}$ via sine rule in both $GMC$ and $BCG$, whence $$\cos^2 \frac{\theta +x}2 = 2\sin x \sin \theta,$$ or, equivalently, $$5 \sin x \sin \theta-\cos x \cos \theta =1.$$ With the change of variable $\tan\frac{x}2\mapsto t$ we get to $$(1−\cos a)t^2−(10\sin a) t+1+\cos a=0,$$ which has the "nice" solution $$t = \frac{(5\pm 2\sqrt 6)\sin \theta}{1-\cos \theta}.$$ Of course, for each $\theta\in(0,180^\circ)$ we only have one solution in $x$. So which sign should we take?

By geometric constraints, $$x < 180^\circ-\theta,$$ and this constraint is satisfied only if we use the minus sign in the above expression. Therefore, the correct solution, for the entire range of $\theta \in (0,180^\circ)$ is

$$\boxed{x_{max} = 2\arctan\left[\frac{(5- 2\sqrt 6)\sin \theta}{1-\cos \theta}\right]}.$$

---

Later edit

A long comment to the comment below by Blue: as he pointed out the relation above can be written as $$\tan \frac{x}2\tan\frac{\theta}2 = K,\tag{3}\label{3}$$ for some constant $K$ that only depends on the ratio $$R = \sqrt{\frac{\overline{BG}}{\overline{BM}}}=\frac{\overline{CM}}{\overline{CG}}.$$ ($x$ here represents again the maximum value of $\measuredangle GBC$, given $\measuredangle ACG=\theta$.

To see this, consider the figure above, where we added the projection $H$ of $C$ on $BM$. Recall that $\measuredangle HCM = \frac{\theta+x}2$. Therefore we can write \begin{eqnarray} R &=& \frac{\overline{CM}}{\overline{CG}}\\ &=&\frac{\overline{CM}}{\overline{CH}}\cdot \frac{\overline{CH}}{\overline{CG}}\\ &=& \frac{\cos\left(\frac{x}2-\frac{\theta}2\right)}{\cos\left(\frac{x}2+\frac{\theta}2\right)}. \end{eqnarray} Note that this expression does not change if $H$ (unlike what is shown in the figure) lies between $G$ and $M$. Using addition formulas and dividing numerator and denominator by $\cos \frac{x}2 \cos\frac{\theta}2$ we obtain $$\frac{1+\tan\frac{x}2\tan\frac{\theta}2}{1-\tan\frac{x}2\tan\frac{\theta}2}=R,$$ i.e. $$K=\tan\frac{x}2\tan\frac{\theta}2 = \frac{R-1}{R+1}.$$ In the case proposed by OP, $R =\sqrt{\frac{3}{2}}$, which yields, as expected, $K = 5-2\sqrt 6$.

---

An even later edit

Since the problem seems to attract many, I wish to add another insight on the role played by equation \eqref{3}. Consider the figure below, where we added $R$ (the center of circle $\gamma$), $Q$ (the midpoint of $GM$), and $P$ (the intersection between $\gamma$ and the perpendicular bisector of $GM$), so that $CP$ bisects $\angle MCG$. Since we know that $\measuredangle MCH = \frac{\theta + x}2$, we have that $$\measuredangle KCH = \measuredangle KPQ = \frac{x}2,$$ and that $CP \perp BN$, where $BN$ is the bisector of $\angle GBC$. As a result, triangle $BKC$ is isosceles and $\overline{BK} = \overline{BC} = \sqrt 6$, independently of $\theta$.

Therefore, for every $\theta$, $\overline{KQ} = \frac{5}2-\sqrt 6$. We have, then \begin{eqnarray} \frac{5}2-\sqrt 6 &=& \overline{KQ}\\ &=& \overline{PQ}\ \tan\frac{x}2\\ &=&\left(\overline{PT}-\overline{QT}\right)\ \tan\frac{x}2\\ &=&\left(\overline{GT}-\overline{QT}\right)\ \tan\frac{x}2\\ &=&\frac12 \left(\frac1{\sin\theta}-\frac{\cos\theta}{\sin\theta}\right)\ \tan\frac{x}2\\ &=& \frac1{2}\tan\frac{\theta}2 \tan\frac{x}2 , \end{eqnarray} yielding again \eqref{3}.

---

Yet another edit

I think one of the most beautiful configurations is the one in which $x$ and $\theta$ are equal.

With the constraint $0< x < 180^\circ$, from \eqref{3} we obtain, in this case, $$\tan\frac{x}{2} = \sqrt{5-2\sqrt 6} = \sqrt3-\sqrt2,$$ i.e. $$\tan x = \frac{1}{\sqrt2}.$$

This is a right-angled triangle with sides proportional to $\overline{BC} = 1$, $\overline{AC} = \sqrt 2$, and $\overline{AB} = \sqrt3$, as depicted in the figure above, where the red markers show the angle whose measure corresponds to $x = \theta = \arctan\frac1{\sqrt2}$.

This is, by the way, the only right-angled triangle having two perpendicular medians. In fact, if we only know this fact and set $\overline{BC} = 1$, by Thales Theorem we have that the projections of the sides $BG$ and $CG$ on $BC$ have measures $\frac23$ and $\frac13$ respectively. By the Geometric Mean Theorem applied to $BCG$, we get that the distance of $G$ from $BC$ is equal to $\sqrt{\frac13\cdot \frac{2}3} =\frac{\sqrt 2}{3}$. We can now apply Thales Theorem "horizontally" and obtain $\overline{AC} = 3\cdot \frac{\sqrt2}3 = \sqrt2$, as desired.

For a geometric approach, let circle with radius $BC$ intersect medians $CD$, $BF$ extended at $H$, $K$. Extend $CA$ to $J$, and let $JK$ and $HC$ extended meet at $L$.

Since $\angle ACG$ is exterior to triangle $CJL$, then$$\angle ACG=25^o=\angle JLC+\angle CJL$$and$$\angle CJL<25^o$$But$$\angle CJL=\frac{1}{2}\angle CBK$$Therefore$$\angle CBK<50^o$$

Fix cevian $CGF$. Point $A$ must vary along a line $\ell$ that forms a $25^\circ$ angle with $CG$ at $C$. This uniquely determines $B$ by reflection about $F$, so $B$ varies along a line $\ell'$ parallel to $\ell$ which passes through $C'$ (the reflection of $C$ about $F$).

Therefore, $\angle CBG$ is maximized when $(CBG)$ is tangent to $\ell$' (see here). Let $\alpha=\angle GCB$. It follows that $\angle GBC'=\alpha$, so $\alpha=\frac{155^\circ-x}{2}$. By Ratio Lemma on $\triangle CBC'$ with cevian $BG$, then on $\triangle ACB$ with cevian $CF$, we have $$\frac{1}{2}=\frac{\sin x}{\sin\alpha}\frac{BC}{BC'}=\frac{\sin x}{\sin\alpha}\frac{BC}{AC}=\frac{\sin x}{\sin\alpha}\frac{\sin 25^\circ}{\sin\alpha}$$ Note that $f(x)=\frac{\sin x\sin 25^\circ}{\sin^2\left(\frac{155^\circ-x}{2}\right)}$ is increasing when $x<\frac{\pi}{2}$ since the numerator is increasing and the denominator is decreasing. We evaluate $f(49^\circ)\approx 0.50007$. Thus, the solution to $f(x)=\frac{1}{2}$ must satisfy $x<49^\circ$.

Here is a geometric solution that explicitly finds the optimal value of $x$.

The setup is the same as my other answer—we know that $A$ varies along some line $\ell$, and by reflection across $F$, point $B$ varies along some line $\ell'$. The optimal angle occurs when $(GCF)$ is tangent to line $\ell'$.

By angle chasing, $\angle CDG=\angle DBC'=\angle C'CB=\angle CC'A$, so $C'ADG$ is cyclic (call its circumcircle $\omega$). Power of a Point from $C$ to $\omega$ gives $$CD\cdot CA=CG\cdot CC'\implies 2CD^2=3CG^2,$$ so $\frac{CD}{CG}=\frac{\sqrt{3}}{\sqrt{2}}$. Multiplying by $2$ gives the $\sqrt{6}:1$ ratio found in @Blue's answer, which is independent of $\theta$.

Now for the formula for $x$. More angle chasing gives $\angle DFG=\angle AC'C=\angle HDG$ (black angles) and $\angle FDG=x=\angle DHG$ (red angles), so $\triangle HDG\sim\triangle DFG$. Thus, \begin{equation} \frac{HG}{DG}=\frac{DG}{FG}\implies \frac{2DG^2}{4FG^2}=\frac{HG}{2FG}\implies \left(\frac{\sqrt{2}DG}{CG}\right)^2=\frac{HG}{CG}.\tag{*} \end{equation}

Law of Cosines on $\triangle CDG$ gives $DG=5-2\sqrt{6}\cos\theta$ when $CG$ is normalized to $\sqrt{2}$. The LHS of (*) becomes $5-2\sqrt{6}\cos\theta$. Law of Sines on $\triangle HCG$ yields $\frac{HG}{CG}=\frac{\sin \theta}{\sin x}$. Using these relations, we rewrite the lengths in (*) to angles: $$\sin x=\frac{\sin\theta}{5-2\sqrt{6}\cos\theta}$$ When $\theta=25^\circ$, the optimal value of $x$ is $\approx 48.995$, which is less than $49^\circ$.

I give here a distinct solution of the precedent ones.

We take the median $CC'$ of the given triangle $\triangle{ABC}$ on the $x$-axis, $C=(0,0),G=(2,0)$ and $C'=(3,0)$. The locus of points $B$ from which angle $\angle{CBG}=49^{\circ}$ is given by the equation $$x^2+y^2-2x-2y\cot(49^{\circ})=0;\space\space y\ge0\hspace2cm(1)$$ This is not other than the locus of inscribed angles of $49^{\circ}$ respect to the chord $\overline{CG}$ in the circle of radius $\sqrt{1+\cot^2(49^{\circ})}$ and center $(1,\cot(49^{\circ})$.

Further we have as given condition the line $y=-\tan(25^{\circ})x$

So the vertex $B$ should be in $(1)$; let $B=(a,b)$ which, because of median $CC'$, gives immediately vertex $A=(6-a,-b)$ which give $\tan(25^{\circ})=\dfrac{b}{6-a}$ which in turn implies that point $B$ is in the line $$y=\tan(25^{\circ})(6-x)\hspace2cm(2)$$ Plotting line $(2)$, it seems to be tangent to the curve $(1)$ in a point, $P$, defined by the intersection of line $(2)$ with its perpendicular $y-\cot(49^{\circ})=\cot(25^{\circ})(x-1)$ passing through the center $(1,\cot(49^{\circ})$ . Calculation gives $P=(1.56007,2.07037)$.

Now if $P=B$ as it seems, then angle $X$ of the problem can be equal to $49^{\circ}$ which would be contrary to the given statement. Putting $P$ in curve $(1)$, we have $$\cot(49^{\circ})=\frac{1.56007^2+2.07037^2-2\times1.566007}{2\times 2.07037}=0.8694364634$$ and $$\cot^{-1}(0.8694364634)=48.9951140793^{\circ}$$ Thus $P\ne B$ which proves that $X\lt49^{\circ}$.

QUESTIONS.-We see that for all angle $X$ such that it is possible the desired configuration for triangle $\triangle{ABC}$, there are just two possibilities, $B_1$ and $B_2$, for the point $B$.

Do the two possible vertices give two different configurations, or is only one valid? Or are two sometimes valid? This can be verified by equating $\overline{BG}=2\overline{GB'}$ and $\overline{BC'}=\overline{C'A}$. For small values of angle $X$, it is not obvious that this holds true, but geometric intuition can be misleading