Let ABC be a triangle with /ABC < /ACB < 90°.
Let D be a point on side BC such that |AD| = |AC|.
The incircle of triangle ABC is tangent to sides
AC and BC at points P and Q respectively. Let J
be the incenter of triangle ABD.

Prove that line PQ bisects line segment CJ.

Although Broll identifies the mid-point of line JC as a key point, we still need to prove that PQ passes through that point. I think that I’ve done it (at last), but perhaps not in the most efficient way..

Let O be the intersection of PQ and JC. Let R and S be the feet of perpendiculars from J and O, respectively, to BC. Let E be a point on BR such that |ER| = |DR|.
Let a, b and c be the sides of triangle ABC.

/OQS = 90^o - C/2           (using the isosceles triangle PQS)

/JER  =  /JDR  =  (/ADB)/2  =  0.5(180^o - C)  =  90^o - C/2  =  /OQS

So triangles JEC and OQC are similar and:      |JC|/|OC| = |EC|/|QC|      (1)

Semi-perimeter of triangle ABD:              s’ = (b + c + |BD|)/2

Therefore             |RD|              =  s’ - c
                                    =  (b - c + |BD|)/2

which gives        |BD| - 2|RD| = c - b       i.e.      |BE| = c - b

Thus                    |EC|   =  a - |BE|
                                    =  a + b - c                                                  (2)

Using (1) and (2):          |JC|/|OC|  =  |EC|/|QC|  =  (a + b - c)/(s - c)
where s is the semi-perimeter of ABC.

Thus  |JC|/|OC|  =  2(s - c)/(s - c)  =  2  proving that PQ bisects JC at O.

Posted by Harry on 2010-10-27 23:41:47