Find all secants to that triangle with same length as PQ and also dividing the perimeter in two.

Choose any point P on AB, and construct, if possible, the point Q on AC with
AP + AQ = (AB + BC + AC)/2 = s, hence AQ = s - AP.

The reflection of PQ through the angle bisector of A gives another solution,
if P' and Q' are both inside the sides AB and AC.

Otherwise, we have to search solutions, that is chords with length PQ, in angles B and C instead of A.

In the above applet, point P is restricted in the suitable part of AB,
that is AQ = s - AP ≤ AC,
then AP ≥ s - AC

If the reflection of PQ through the angle bisector of A is inside segments AB, AC, it is also drawn.

Now comes the true question : to construct if any, segments P"Q" = PQ inside angles B and C.

For instance inside angle C.

Let's construct from PQ a point N with <)PNQ = C and NP + NQ = s

This is the intersection of a circle (from which we see PQ under angle C) and an ellipse (NP + NQ = s) with foci P and Q,
which doesn't seem so easy...

Review the construction of ellipse points from a circular directrix.

The current point N on ellipse is constructed from a current point M on the circular directrix,
centered in Q with radius s :

The ray QM and the perpendicular bisector of PM intersect in N.

Triangle PNM is isosceles and the perpendicular bisector of PM is also angle bisector of PNM.

We deduce an interesting property : <)PMQ = PNQ/2

The source point M for searched point N is then the intersection of the circular directrix with the circle
from where we see PQ under angle C/2.

The triangle PNQ is then copied in P"CQ" inside angle C, if possible,
that is if P" and Q" are inside segments AC and BC.

The second solution inside angle C comes from the other intersection M', or just by a reflection of the first candidate through the angle bisector of C.

The construction of the circle from which we see PQ under angle C/2 is not detailed in the applet.
This elementar construction may be done as follows :

Copy angle C along PQ, at Q, then divide by 2 and draw the perpendicular in Q to that angle bisector.

It intersects the perpendicular bisector of PQ in center of the searched circle.