Two triangles have an equal angle, inradius, circumradius. Are these two triangles necessarily congruent?
Let $A_1,A_2$ be the areas if the triangles respectively, $s_1,s_2$ be the semi perimeter respectively, and $a_1,b_1,c_1$ and $a_2,b_2,c_2$ be the side lengths of the two triangles respectively. Then, $$A_1/s_1 = A_2/s_2$$ Also, $$a_1 b_1 c_1/ 4A_1 = a_2 b_2 c_2/ 4A_2$$ Also, as the circumeradius is equal, we get $a_1=a_2$. How do we proceed?
Euler's theorem says:
$OI^2=d^2=R(R-2r)$
where R and r are circuncircle and in-circle radii respectively. We may also use this theorem that says:
If segment connecting centers of two circles OI satisfy this relation ; $OI^2=R(R-2r)$, then we can inscribe infinite triangles in radius R having inscribed circle with radius r. As can be seen in figure, these triangles are not necessarily congruent. Only triangle which is the mirror of original triangle about a diameter of circumcircle perpendicular on IO at O, is congruent with original one, as shown in figure b. also we have:
$R=\frac {abc}{4A}$
$r=\frac A s$
Mulyiplying these we get:
$R\cdot r=\frac{abc}{2(a+b+c)}$
$OI^2=d^2=R(R-2r)=R^2-2R\cdot r=R^2-\frac{abc}{a+b+c}$
$\Rightarrow \frac{abc}{a+b+c}=2R\cdot r=R^2-d^2$
this is Diophantine equation with unknown a, b and c, which due to the geometric theorem can have infinitely many solution, which are not necessarily equal.
Update: It seems the statement is changed. The new one includes one angle equal in two triangles. This is sufficient condition for congruence of two triangles; that is; if two triangles have equal inscribed and circumscribed circles, for their congruence it is sufficient that one of their angles be equal.
that is the answer is Yes, they are necessarily equal(figure b).
Proof: consider figure b. we have:
$BI=\frac r{sin \hat{ B}/2}$
$OI=d=\sqrt{R(R-2r)}$
$BO=R$
that means triangle BIO is unique for certain value of $\angle ABC$, r and R which results in triangle ABC is unique, in other words having these parameters you can only construct one triangle in which OB and IB meet on circumcircle.