Triangle $ABC$ is isosceles with $AB = AC$. $P$ is a variable point on $AB$, and $Q$ is a variable point on $AC$, so that $BP = AQ$. Let $O$ be the midpoint of $PQ$. Prove that $d(O, BC)$ is constant, and determine the locus of point $O$.
I feel that it should be solved with the help of center lines, and that the locus of point $O$ will be the centroid of a triangle $ABC$, but I have not been able to prove it.
I need some hints. How can I start?
On
Let P be a point on AB of the isosceles triangle ABC.
Through P draw the red line parallel to AC. It will cut BC at X.
Through X draw the red dotted line parallel to AB, it will cut AC at some point T, say.
Since APXT is a //gram, AT = PX = PB = AQ. This means T is actually Q. Also, if $O_1$ is the midpoint of the diagonal PQ, it is also the midpoint of the diagonal AX.
Next, starting with some other P’ on AB, and going through the similar process as the above, we get the corresponding pint $O_2$.
Applying the midpoint theorem to triangle AXX’, we conclude that $O_1O_2 // XX’$. Result follows.
In the limiting case that when P moves to the midpoint of AB, that midpoint is $O_n$ for some n. The same thing happens to Q. We can say that the locus of O is the mid line of the isosceles triangle ABC.
First we translate BP to find the Q on AC. For this we take the mid point of AB, M, and reflect P about it to get point P'. Now we draw a circle center on A and radius AP', it intersect AC at Q(why?). In triangle PP'Q, MO is parallel with P'Q and in triangle ABC P'Q is parallel with BC, therefore $MO||BC$, this property holds for any position of Point P which is the start point of construction. In this way the locus of point O is a line that connects the mid points of legs AB and AC.