Find locus of Q of an equalateral triangle APQ with A fixed and P moving along a line

Question

ABC is an equilateral triangle with vertex A fixed and B moving in a given straight line. Find the locus of C

I drew some figures and realised that locus is a straight line and the part left is to prove

Now, consider a equilateral triangle ABC and consider a line $l$ passing through B. Let us take any point P on line $l$,join PQ and AQ. As per question this new triangle APQ must be equilateral.This is what I have to prove

After thinking for some time and doing some work, i realised that to prove the statement it suffice to prove that $ABP \cong ACQ$. I am struggling to prove it.I thought about in all ways but still i am unable to prove it (try to prove it by only pure elementary method ) EDIT: here angle APQ = 60.

Answer 1

In general, it can be shown that the locus of the vertex Q is

$$(\sqrt3 -\tan\theta)x-(\sqrt3\tan\theta +1)y = a(\sqrt3+\tan\theta)\tag 1$$

where $\theta$ is the angle between the lines BQ and BP and the $a$ is the side length of the equilateral triangle ABC. Note that the locus (1) is a straight line. However, it does not coincide with the line BQ, as assumed in the question. Only in the special case of $\theta = 60^\circ$, it is so as the locus (1) reduces to

$$ y = -\frac{\sqrt3}2a$$

Assume then $\theta = 60$. It can be shown geometrically that the triangle APQ is equilateral. Extend AB to D and construct the equilateral triangle BDP. Given that $\angle APD = \angle BPQ$, $\angle ADP = \angle QBP$, as well as BP = DP, the triangles ADP and QBP are congruent. Thus, AP = QP = AQ.

Answer 2

Observe a rotation around $A$ for $60^{\circ}$.

This map takes $B$ to $C$, so when $B$ describes a line $\ell$ then $C$ describes a line $\ell'$ which a picture of $\ell$ under this rotation.