Construct two points B0 and B1 on line AB with B0
on ray AB and |AB0| = |AB1| = |BC|/2. Construct
points C0 and C1 similarly on line AC.
Let Γij denote the circle that is tangent to line
AB at point Bi and tangent to line AC at point Cj.
The number of solutions depends on the location of
point D. Let ray AD intersect Γ00 at points E and F
where 0 < |AE| < |AF|.
Case I: |AD| > |AE| - (Two solutions)
There exist two lines through D tangent to Γij (i≠j).
Construct the one where the point of tangency T
lies on the smaller arc BiCj. Proof:
|AB'| + |B'C'| + |C'A|
= |AB'| + ( |B'T| + |TC'| ) + |C'A|
= |AB'| + ( |B'Bi| + |CjC'| ) + |C'A|
= ( |AB'| + |B'Bi| ) + ( |CjC'| + |C'A| )
= |ABi| + |ACj|
= |BC|
Case II: |AD| = |AE| - (Three solutions)
In addition to the two solutions of Case I, the
line through D and tangent to Γ00 is also a
solution. Proof is similar to Case I.
Case III: 0 < |AD| < |AE| - (Four solutions)
In addition to the two solutions of Case I, the
two lines through D and tangent to Γ00 are also
solutions. Proof is similar to Case I.
QED
|
