TITLE: A triangle where two of its sides are diameters of circles
QUESTION [2 upvotes]: Given an acute triangle $\triangle XYZ$ and two circles with diameters $XY$ and $YZ$, the circles intersect at two points. One of the intersection points is obviously at $Y$ by construction. 
Is the other intersection point $W$ inside, outside, or on $\triangle XYZ$? How to prove it?

REPLY [0 votes]: Consider $\mathbb R^2=\mathbb C$. By a Moebius transformation
$F$ interchange point $Y$ with $\infty$. Such a Moebius maps sends:
Lines not through $Y$ to circles through $Y$.
Lines through $Y$ to lines through $Y$.
Circles through $Y$ to lines not through $Y$.
Circles not through $Y$ to circles through $Y$
So:
The line $L$ between $X,Z$ is mapped to the circle $C=F(L)$ trough $F(X),F(Z),Y$.
The circle $C_1$ whose diameter is $ZY$ is mapped to a line $A_1$ through $F(Z),Y$
The circle $C_2$ whose diameter is $X,Y$ is mapped to a line $A_2$ through $F(X),Y$
The line $L_1$ through $ZY$ is mapped to a line $B_1$ through $F(Z),Y$
The line $L_2$ through $XY$ is mapped to a line $B_2$ through $F(X),Y$
Since $C_1$ and $L_1$ are perpendicular and moebius maps are conformal, then $A_1$ and $B_1$ are perperndicular at $F(Z)$
Same reasoning for $A_2$ and $B_2$ that are then perpendicular at $F(X)$.
It follows that $P=A_1\cap A_2$ is the other point of the diameter of $C$ opposite to $Y$, and that triangles $\Delta YF(Z)P$ and $\Delta YF(X)P$ are the two rectangle triangles inscribed in $C$ with diameter $YP$.
Since $C_1$ and $C_2$ intersects at $W$, then $A_1$ and $A_2$ intersect at $F(W)$.
So $P=F(W)$, hence $F(W)\in C=F(L)$ and so $W\in L$.
(A picture would help...)
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