Prove that the triangles $ABC$ and $AB^{'}C^{'}$ have the same incentre.

Question

The question is as follows if $ABC$ is a triangle, with $AD$ as the internal angle bisector of $\angle A$ with $D$ at $BC$ and $B^{'}, C^{'}$ are reflection of points $B$ and $C$ in $AD$. Show that triangles $ABC$ and $AB^{'}C^{'}$ have the same incentre. This was the exact question language,

Now intuitively here was where I could not satisfy myself, how come two triangles which only shared a vertex $A$ above have the same incentre ?all I could find in this problem was we have $C^{'}B^{'} = CB$,(because notice that here $B$ and $B^{'}$ lie at a same distance from $AD$ as $B^{'}$ is a reflection of $B$ only,same with $C$ and $C^{'}$) which also gives us $BD = DB^{'}$ and $CD = DC^{'}$ , all I claim here could be proved if we join $BB^{'}$ and $CC^{'}$ which pass through $AD$ and is perpendicular to it and also these 2 pair of points($B, B^{'}$) & ($C, C^{'}$) lie equidistantly from $AD$, by which I could say triangle $BDC^{'}$ is congruent to $B^{'}DC$. I didn't find anything more substantial.

So here is my question from the figure can we ever have the incentre of triangle $ABC$ to be the incentre of $AB^{'}C^{'}$ for incentre we need a centre of a circle inside the triangle which touches all the 3 sides of it which so how come another circle of a different triangle have the same incentre ? Although I could not disprove it concretely it just runs counterintuitive to me, so do I miss something Or did I interpret the question wrongly ?

Answer 1

Your figure looks a bit off.

Triangle $AB^\prime C^\prime$ is very closely related to triangle $ABC$. In particular, one is the reflection of the other in the line $AD$. Since the incenter of $ABC$ lies on $AD$, reflecting it in $AD$ leaves it unchanged.

Answer 2

The diagram that is shown is misleading, because if $AD$ is the angle bisector of $\angle BAC$, then $$\angle BAD \cong \angle CAD,$$ consequently the reflection $B'$ of $B$ in $AD$ will be collinear with $AC$, and the reflection $C'$ of $C$ in $AD$ will be collinear with $AB$.

The incenter $I$ of a triangle is the mutual intersection point of the angle bisectors of all interior angles of the triangle. This should be clear once you recognize that the incenter, being equidistant from the three sides, must form pairs of right angled triangles, each pair sharing a common leg (the inradius) and common hypotenuse (the distance from the incenter to a vertex). Therefore, the resulting pairs of right angled triangles are congruent, and their respective angles also congruent; thus the lines joining each vertex to the incenter must be angle bisectors.