[freemind]

In triangle the bisector of intersects at . Consider an arbitrary circle passing through and , so that it is not tangent to or . Let and .
a) Prove that there is a circle so that and are tangent to in and , respectively.
b) Circle intersects lines and in and respectively. Prove that the lengths of and do not depend on the choice of circle .

[Clark Kent]

freemind wrote:

In triangle the bisector of intersects at . Consider an arbitrary circle passing through and , so that it is not tangent to or . Let and .
a) Prove that there is a circle so that and are tangent to in and , respectively.
b) Circle intersects lines and in and respectively. Prove that the lengths of and do not depend on the choice of circle .

Am i wrong?

[pohoatza]

Denote by the intersection point of the perpendiculars in , to the lines , and , respectively. Since the quadrilaterals and are cyclic, we have that . The triangles and are now congruent, and therefore the segments , and are equal. Thus, the circle with center and radius is tangent to the lines , at points , , respectively. This proves a).

Next, notice that the triangles and are congruent (this is because , , and is common), and thus . Similarly, the triangles and are congruent, and hereby . Since , it is suffice to prove that is constant. But this follows from Ptolemy's theorem, applied to the quadrilateral , since

(we know that from the congruence of the triangles and ), and . This proves b).