It is known that three given positive numbers determine a unique triangle with the angle bisectors lengths equal to these numbers. Therefore two triangles are congruent on three angle bisectors.
Immediately following the above conclusion, here's an interesting question.
Are two triangles congruent if the distances from the incenter of the triangle to the three vertices correspond to equal distances?
My attempts：Let the incenter of triangle $ABC$ be point $P$, and $AB=c,AC=b,BC=a$, $AP=x,BP=y,CP=z$, then we can get $$ \begin{cases} x^2=\dfrac {bc(b+c-a)}{a+b+c}\\ y^2=\dfrac {ac(a+c-b)}{a+b+c}\\ z^2=\dfrac {ab(a+b-c)}{a+b+c} \end{cases} \qquad (*) $$ The problem is equivalent to the question of whether $a,b,c$ are unique given positive numbers $x,y,z$ that satisfy the system of equations $(*)$. However,It is difficult for me that no substantial progress was made.
Is there a better way to solve this problem?
Yes, the 3 distances from the incenter to the 3 vertices determine a triangle up to congruence.
Let the given distances from the incircle center $I$ to the 3 triangle vertices $A,B$ and $C$ be $x$,$y$ and $z$, resp. Let the (as of now unknown) incircle radius be $r$. The situation is shown in the picture below, where I've only shown things relevant to side AB, which is touched by the incircle at point D.
We start with noticing $\angle AIB = \angle AID + \angle DIB$. Note that since $D$ always lies between $A$ and $B$, the sum is indeed always the correct way to get the whole angle from the 2 parts.
We have
$$ \cos \angle AID = \frac rx,\; \cos \angle DIB = \frac ry.$$
Since $0° < \angle AID, \angle DIB < 90°$ we finally get
$$\angle AIB = \arccos \frac rx + \arccos \frac ry.$$
The same can be done for the 2 other sides (with corresponding angles at $I$), so we get
$$360° = \angle AIB + \angle BIC + \angle CIA = 2\left(\arccos \frac rx + \arccos \frac ry + \arccos \frac rz\right).$$
The expression on the right hand side can be considered a function of the unknown $r$, and as such it is strictly decreasing as the sum of 3 strictly decreasing $\arccos$ functions. That means $r$ is uniquely determined by $x,y$ and $z$. But with $r$ being able to achieve only 1 possible value, the triangles $\triangle AID$ and $\triangle IDB$ are completely determined, then so is $\triangle AIB$ and by anylogy $\triangle BIC, \triangle CIA$ und thus the whole $\triangle ABC$.