"$AB$ and $CD$ are two fixed straight lines and a variable straight line cuts them at $X$ and $Y$ respectively. The angular bisectors of $\angle AXY$ and $\angle CXY$ meet at $P$. Find the locus of $P$."
First things first, if the question states $\angle CXY$, it is clear that their angle bisectors meet on line $AB$ and thus, $P$ lies on $AB$. I believe the question wanted to state "$\angle AXY$ and $\angle CYX$" instead. So I considered the question to be stated as,
"$AB$ and $CD$ are two fixed straight lines and a variable straight line cuts them at $X$ and $Y$ respectively. The angular bisectors of $\angle AXY$ and $\angle CYX$ meet at $P$. Find the locus of $P$."
My attempt:
It is trivial to note that had $AB$ and $CD$ been parallel, the locus would have been $\angle XPY=90^{\circ}$. Then I tried considering non-parallel lines.
Assume that $\angle AXY≥\angle BXY$, and $\angle CYX≥\angle BYX$ such that $\angle AXY= 2x$ and $\angle BXY= 2y$. Then, if $P$ lies on the side whose sum of interior angles$>180^{\circ}$, clearly $\angle XPY=180-(x+y)$. The same may be done for all combinations of relations of supplementary pairs $(\angle AXY,\angle BXY)$ and $(\angle CXY,\angle ∠DXY)$.
However, this is an EXTREMELY piece-wise definition (and subject to errors) and as a result, I'm not satisfied with it; there must be a way to refine my answer to produce a much more accurate and concise one. In short, I have 2 primary questions:
- If my method of solving is correct, how do I refine my answer?
- If I need to re-start from scratch, how do I start?
Thanks for the help!
Source: Challenge and Thrill of Pre-College Mathematics
*Note: This may be marked a duplicate as I've seen the same question before. However as I couldn't find a satisfactory explanation there, I decided to post my own question.
As you have already considered the scenario where $AB\parallel CD$, i am considering these two straight lines two be non parallel.
Let $AB$ and $CD$ meet at point $Q$. Divide this into two cases.
Case $1:$ We get point $Q$ by extending $AB$ and $CD$ through points $A$ and $C$ respectively.
Then, in $\triangle QXY$, the internal bisectors of $\angle QXY$ and $\angle QYX$ meet at $P$.Hence, $P$ is the incentre of $\triangle QXY$ and thereafter the locus of $P$ is the bisector of $\angle Q$.
Case $2:$ We get point $Q$ by extending $AB$ and $CD$ through points $B$ and $D$ respectively.
Then, in $\triangle QXY$, the external bisectors of $\angle QXY$ and $\angle QYX$ meet at $P$. Hence, $P$ is the excentre of $\triangle QXY$ and thereafter the locus of $P$ is the bisector of $\angle Q$.
Thus, the locus of $P$ is the line which internally bisects the angle formed by the lines $AB$ and $CD$.