kinematics - Why do three particles placed at the vertices of an equilateral triangle, moving along the sides, meet at the centroid?

I just read an example of vectors in my book which is confusing me.

Three particles A,B and C are at the vertices of an equilateral trinagle ABC. Each of the particle moves with constant speed v. A always has its velocity along AB, B along BC and C along CA. They meet each other at the centroid. At any instant, the component of velocity of B along BA is $v\cos60^\circ$.

I don't understand how the particles meet at the centroid and why the component of velocity of B along BA is $v\cos 60^\circ$.

Answer

This question is exactly 11th class's question in Kinematics chapter. It is used in many books. The book in my hands is H.C Verma's Concept of physics and chapter-3 solved example-20.

Your answer: By symmetry points$A(t),B(t),C(t)$ will always make an equilateral triangle. Since the angle b/w $BC$ and $BA$ is always $60^0$ so the component of velocity of $B$ along $BA$ is always $v\cos60^0$.
All the triangles $A_1B_1C_1,A_2B_2C_3$ and $A_nB_nC_n$ are concentric. Hence at the end when $A$,$B$ and $C$ approach each other they form an infinitesmall triangle whose centroid is the same as that of the initial triangle $A_1B_1C_1$ and is forthcoming point of meeting of $A$,$B$ and $C$.