Let D be a point in the plane of ΔABC such that
ray AD intersects the interior of side BC.

Construct a line through D intersecting lines AB and
AC at points B' and C' respectively such that the
perimeter of ΔAB'C' equals |BC|.

Bisect the line BC and call its mid-point M.
Construct the parallelogram AMBP.
Draw the circle, c1, with centre at A and which passes through P.
Denote the intersections of c1 with AP, AB, AC  by Q, R, S respectively.
Construct perpendiculars to AB and AC through R and S respectively.
Let T be the point of intersection of these perpendiculars.
Draw the circle, c2, with centre at T which passes through R and S.
Construct the circle, c3, with DT as diameter, crossing c2 at E.
DE will touch c2 at E and is the required line.

(The line will exist only if D is outside the circle c2, as Jer pointed out.
Also, c2 and c3 will intersect at two points, giving two possible positions
for E and hence two possible positions for the line B'C').

Proof

  Perimeter of AB'C'        =  AB' + B'E  + C'E + AC'
                                    =  AB' + B'R + C'S + AC'                        (tangents equal)
                                    =  AR  +  AS
                                    =  PA  +  AQ                  (equal radii of c1)
                                    =  BC                (oppo. sides of parallelogram)  


Edit: Don't know why all these lines are wrapped round - Sorry!

Edited on May 12, 2013, 3:57 pm

Posted by Harry on 2013-05-12 15:52:16