Rename $Z$ as $F$. Now, consider the polar map with respect to the circle $\omega$ with center $A$ that passes through $B$ and $D$. We know that the pole of $E$ will pass through the points of tangency from $E$ to $\omega$ and will be perpendicular to $AE$ - note that $DF$ does indeed pass through one of these tangency points ($D$) and satisfies the orthogonality requirement; these two requirements uniquely determine the line, so $DF$ is the pole of $E$, i.e. $F$ lies on the pole of $E$ so $E$ must lie on the pole of $F$. As $B$ is a point of tangency from $F$ to $\omega$, the pole of $F$ is thus $BE$ and must be perpendicular to $AF$.

A general enunciation: Let $m,n$ be two lines (which can be parallels or concurrently) and a circle $w=C(I,r)$ which is tangent to the lines $m,n$ in the points $A\in m$, $B\in n$ respectively. Prove that for any points $M\in m$, $N\in n$ the following sentencies are equivalently: $1^{\circ}.\ IM\perp AN\ ;\ \ \ \ 2^{\circ}.\ IN\perp BM\ ;\ \ \ \ 3^{\circ}.\ AM^2+BN^2=MN^2.$ A particular cases. $\blacksquare\ a).\ m\parallel n$. Let $AMNB$ be a trapezoid ($AM\parallel BN$) and the middlepoint $I$ of the side $[AB]$. Prove that $IM\perp AN\Longleftrightarrow IN\perp BM\Longleftrightarrow AM^2+BN^2=MN^2.$ $\blacksquare\ b).\ P\in m\cap n$. Let $ABC$ be a triangle. The its incircle $w=C(I,r)$ touches the its sides in the points $D\in BC$, $E\in CA$, $F\in AB$ and the its exincircle $w_a=C(I_a,r_a)$ touches the its sides in the points $D_1\in BC$, $E_1\in CA$. $F_1\in AB$. I note the intersections $L\in BC\cap EF$ and $L_1\in BC\cap E_1F_1$. Prove that: $IL\perp AD$, $LA^2=LD^2+AE^2$; $I_aL_1\perp AD_1$, $L_1A^2=L_1D_1^2+AE_1^2$. (A remarkable property !)

Silouan, in my notes I see a vector solution(inner product), but I do not remember where I have found it. Do you have the same solution? Babis

Stergiu, read you the my extension of this problem (with three degrees of freedom) ?

@mr Stergiou : Yes my solution is as you said above. There exist another solution with elementary geom in your book for C' of Gymnasium at page 185.

Let $F\in AE\cap DZ, G\in BE\cap AZ$, From $\triangle ADE\implies AD^2=AF\cdot AE$, but $AB=AD$ ans so $AB$ is tangent to the circle $\odot BEF\implies\angle ABF=\angle AEB=\angle AEG$. $ABZF$ being cyclic, $\angle ABF=\angle AZF=\angle GZF$, consequently $\angle GZF=\angle GEF$ and $EFGZ$ is cyclic, hence $\angle EGZ=\angle EFZ=90^\circ$, done. Remark: $BF, DG, AC$ are concurrent, prove! My best wishes for a Merry Christmas and a Happy, and Prosperous New Year 2015 everybody! Best regards, sunken rock