This is a late attempt to contribute to an interesting discussion I only recently became aware of. I hope the question is still alive.

**If only one angle in the given triangle is greater than $60^o$, then the greatest inscribable equilateral triangle shares a vertex with this greatest angle and has its base on a portion of the opposite side, as in the OP's first figure.**

Presuming this as agreed on, and also that the greatest equilateral triangle must share at least one vertex with the given triangle, I consider only the case where *two* angles are greater than $60^o$,

1) showing that the greatest equilateral triangle does not always share a vertex with the largest angle in the given triangle, while

2) revealing the conditions under which it *does*, and

3) the method for finding the greatest equilateral triangle when it *does not*.

**Suppose two angles of the given triangle are greater than $60^o$ and equal.**

In isosceles triangle $ABD$, the greatest equilateral triangle has $AB$ as base when the base angles of $ABD$ are greater than $60^o$ but less than $80^o$, i.e.$$\frac{\pi}{3}<\angle DAB=\angle DBA<\frac{4\pi}{9}$$ For if on $AD$ we make $AE=AB$ and construct equilateral triangle $AEF$, vertex $F$ lies outside of triangle $ABD$ unless$$\angle ABF\le\angle ABD$$If $\angle ABF=\angle ABD$, as in the figure above, then $\triangle ABD$ and $\triangle BFA$ similar isosceles triangles,

and since$$\angle BAF=180^o-2\angle ABF$$and also$$\angle ABF=\angle BAD=\angle BAF+60^o$$then$$180^o-2\angle ABF=\angle ABF-60^o$$making$$3\angle ABF=240^o$$and $$\angle ABF=80^o$$

Thus for a base angle of $80^o$, a greatest equilateral triangle can be constructed on a side of the isosceles triangle as well as on its base.
For base angles greater than $80^o$, however, i.e. when$$\angle ABF<\angle ABD$$ point $F$ lies within $\triangle ABD$, and the greatest equilateral triangle is not on the base of the isosceles triangle but on its side.

Accordingly, in the second figure (vertex $D$ of isosceles $\triangle ABD$ is out of view on the right), $F$ is within $\triangle ABD$. The maximum equilateral triangle $AHG$ is constructed by extending $AF$ to $H$ on $BD$, drawing a circle with radius $AH$, cutting $AD$ at $G$, and joining $HG$.

As $D$ becomes more and more remote, $AD$ and $BD$ approach parallelism, and $\triangle AHG$ approaches as a limit an equilateral triangle with height $AB$. If $AB=1$, the side of this limiting greatest equilateral triangle is $\frac{2}{\sqrt 3}>1$.

**If two angles of the given triangle are greater than $60^o$ but not equal, then the greatest equilateral triangle is either on the side between the vertices of these two angles, or shares a vertex only with the lesser of the two angles.**

In the third figure, triangle $ABD$ has angles at $A$ and $B$ both greater than $60^o$, with $$\angle DAB>\angle DBA$$

If $\angle CBD<\angle ADB$, then point $F$ of equilateral triangle $AEF$ lies outside of $\triangle ABD$, so that the greatest equilateral triangle is either $\triangle ABC$ or lies on $BD$ with a vertex at $B$.

But if, with $AD$ fixed in position, we move $D$ to the right, thereby increasing $\angle CBD$ and decreasing $\angle ADB$, until $\angle CBD=\angle ADB$, then point $F$ lies on $BD$ and $\triangle AEF$ is an equilateral triangle in $\triangle ABD$ equal to $\triangle ABC$.

Proof:

Since$$\angle EAC=60^o-\angle CAF$$and$$\angle CBD=\angle CBF=\frac{1}{2}\angle CAF$$then$$\angle EAC=60^o-2\angle CBD$$And since$$\angle ADB=180^o-2\cdot 60^o-\angle EAC-\angle CBD$$therefore$$\angle ADB=60^o-60^o+2\cdot \angle CBD-\angle CBD=\angle CBD$$

And so conversely, if $\angle CBD=\angle ADB$, then point $F$ lies on $BD$.

Now on $BD$ make $BG=AB$, and construct equilateral triangle $BGH$ equal to $\triangle ABC$.

**Then vertex $H$ lies within $\triangle ABD$.**

For suppose $H$ lies on $AD$:

Then since $\angle ABH=\angle CBG$, by SAS we have$$\triangle ABH\cong \triangle CBG$$But$$\angle CBG=\angle ADB$$Therefore$$\triangle ABH\sim\triangle ADB$$But $\triangle ABH$ is isosceles. Therefore, $\triangle ADB$ is isosceles and$$\angle DAB=\angle DBA$$which is false.

Therefore, $H$ lies within $\triangle ABD$, and, by the method already indicated in the second figure, an equilateral triangle greater than $\triangle BGH$ can be constructed on $BD$, sharing a vertex with the angle at $B$, which is *less* than the angle at $A$.

Although this case by itself, i.e. where $F$ lies on $BD$, proves that the conjecture is not always true, it appears that when two angles are greater than $60^o$ the conjecture is in fact universally false: when $D$ is remote enough so that $H$ and $F$ both lie within $\triangle ADB$, the greatest equilateral triangle inscribable on $BD$ with vertex at lesser angle $B$ is always greater than that inscribable on $AD$ with vertex at greater angle $A$.

In the daunting fourth figure, again$$\angle DAB>\angle DBA$$but $D$ is far enough out to the right in triangle $ADB$ so that $F$ and $H$ both lie within $\triangle ADB$. Extend $AF$ and $BH$ to cut $BD$ and $AD$ at $I$ and $J$, complete the larger equilateral triangles $ALI$ and $BJK$, and join $IJ$ and $KL$.

Granting the apparent concyclicity of $J,A,B,I$ and $J,L,K,I$ (I need a proof), then since$$\angle JLK=\pi-\angle JIK$$then$$\angle JLK=\angle JIB$$But$$\angle JIB=\pi-\angle JAB$$Therefore$$\angle JLK=\pi-\angle JAB$$so that$$LK\parallel AB$$Hence $LK$ cuts the sides of $\triangle ADB$ proportionally, making$$\frac{AD}{DB}=\frac{AL}{BK}$$And since$$AD<BD$$therefore$$AL<BK$$and equilateral triangle $ALI$ sharing vertex $A$ of the greater angle $DAB$ is less than equilateral triangle $BJK$ sharing vertex $B$ of the lesser angle $DBA$.

To conclude: apparently when the given triangle has two angles greater than $60^o$, if the greatest equilateral triangle shares a vertex with the greater of the two angles, it also shares a vertex with the lesser, but not *vice-versa*, i.e. the greatest equilateral triangle has as base either the side between the two angles greater than $60^o$, or a portion of the side opposite the greatest angle and shares a vertex with the *lesser* of the two angles greater than $60^o$.