TITLE: Triangle problem
QUESTION [10 upvotes]: I have got one simple task to prove: 
We have got a triangle $\triangle XYZ$.
Then we create points $A,B,C$ on $XY, YZ, ZX$ respectively, such that $XA = AB = BZ$ and $CZ = AY = AC$.
How to prove that $XY = \frac{XZ + YZ}{2}$?

REPLY [4 votes]: I refer to the following figure:

The points $B$ and $C$ have to lie on the perpendicular bisector of $ZA$. The angles $\angle(AZX)$ and $\angle(AZY)$ will be fixed later; call them $\lambda-\delta$ and $\lambda+\delta$. Then the angles at $X$ and $Y$ will be $\lambda'-\epsilon$ and $\lambda'+\epsilon$ with $\lambda':={\pi\over2}-\lambda$ and some $\epsilon$. (The idea is that $|\delta|$, $|\epsilon|\ll 1$.)
The conditions $ZC=CA$ and $ZB=BA$ are already taken care of. The conditions $ZB=AX=:x$ and $ZC=AY=:y$ lead by means of the sine theorem to
$$x={1\over\cos(\lambda+\delta)}={2\sin(\lambda-\delta)\over\sin(\lambda'-\epsilon)}\ ,\qquad y={1\over\cos(\lambda-\delta)}={2\sin(\lambda+\delta)\over\sin(\lambda'+\epsilon)}$$
or
$$\sin(\lambda'-\epsilon)=2\sin(\lambda-\delta)\cos(\lambda+\delta)\ ,\qquad
\sin(\lambda'+\epsilon)=2\sin(\lambda+\delta)\cos(\lambda-\delta)\ .\qquad(1)$$
Adding the two equations $(1)$ gives $2\sin\lambda'\cos\epsilon=2\sin(2\lambda)$ or
$$\sin\lambda ={1\over2}\cos\epsilon\ .\qquad(2)$$
(Subtracting the two equations $(1)$ gives $2\cos\lambda'\sin\epsilon=2\sin2\delta$, so that together with $(2)$ we obtain
$$\sin(2\delta)={1\over4}\sin(2\epsilon)\ .$$ This shows that there is in fact a one-parameter family of such configurations. The value $\epsilon=0$ corresponds to an equilateral triangle $XYZ$.)
We now compute
$$\eqalign{XY=x+y&={2\sin(\lambda-\delta)\over\cos(\lambda+\epsilon)}+{2\sin(\lambda+\delta)\over\cos(\lambda-\epsilon)}\cr &={2\sin(\lambda-\delta)\cos(\lambda-\epsilon)+2\sin(\lambda+\delta)\cos(\lambda+\epsilon)\over
\cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}\cr
&={\sin(2\lambda-\delta-\epsilon)+\sin(2\lambda+\delta+\epsilon)\over
\cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}\cr
&={2\cos(\delta+\epsilon)\over
\cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}\sin(2\lambda)\ .\cr}$$
On the other hand, again by the sine theorem we have
$$ZY=2{\cos(\delta+\epsilon)\over\cos(\lambda-\epsilon)}\ ,\qquad
ZX=2{\cos(\delta+\epsilon)\over\cos(\lambda+\epsilon)}\ ,$$
so that
$${ZX+ZY\over2}=\cos(\delta+\epsilon){\cos(\lambda+\epsilon)+\cos(\lambda-\epsilon)\over\cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}={2\cos(\delta+\epsilon)\over\cos(\lambda+\epsilon)\cos(\lambda-\epsilon)}\cos\lambda\cos\epsilon\ .$$
From $(2)$ follows $\ \cos\lambda\cos\epsilon=\sin(2\lambda)$, so that we indeed obtain
$${ZX+ZY\over2}=XY\ .$$