Obviously $BAC = BCZ$. However, $p$ is parallel to $t$, so $EDC = DCZ$. Consequently, $BAC = EDC$, so $ABDE$ is circumscribable.

(p,AB)=(AC,AB) hence p is antiparallel to BC. So any line DE parallel to p is also antiparallel to BC, hence ABDE cyclic.

Let $O$ be the circumcenter of $\triangle ABC$ and $T$ be a point on $t$ such that $T$ and $B$ are on opposite sides of $OC$. Since line $t$ is tangent to the circumcircle of $ABC$, $$\measuredangle ABC=\measuredangle ACT=\measuredangle CED=\measuredangle AED$$so $ABDE$ is cyclic, as desried. $\blacksquare$

The following Lemma helps: Lemma:Let $\Delta ABC$ be inscribed in a circle with center $O$ and Let line $\mathcal{L}$ be tangent at $C$, and let two points on either side of $C$ be $X$ and $Y$ that lie on line $\mathcal{L}$, such, that $X$ is closer to $A$ than $B$ and $Y$ is closer to $B$ than $A$, Then, $\angle B=\angle ACX$ and $\angle A=\angle BAY$

If $S$ is a point in the tangent through $C$ close to $E$, we have, $\angle BDE=\angle BCS=\angle SCA +\angle C=\angle B +\angle C=180-\angle A$

Isn't this problem too easy for JBMO It seemed to me really simple!

It is a junior problem.

What do you mean @Akihi ??? I am 13 years old!

Hmm it is too easy for shortlist of olympiad.

Let $F$ denote the altitude from point $E$ to line $t$. Then we have \[ \angle{BDE}=\angle{BCF}=\angle{ACF}+\angle{BCA}=\angle{ABC}+\angle{BCA}=180-\angle{BAC} \]Thus we know that $ABDE$ is cyclic.

Let X be a point on tangent of C. ∠ACZ = ∠ABC and ∠ACZ = ∠DEC ---> ABDC is cyclic. dude this problem is a joke how can such thing be in shortlist

Since ${t}$ is tangent to $\odot$$(ABC)$ we get that: $\angle$$(t,AC)$$=\angle ABC=\beta$ ${t}$ $\parallel$ ${p}$ $\implies$ $\angle$$(t,AC)$$=\angle DEC=\beta$ $\angle DEC + \angle AED=180$ $\angle AED=180-\beta$ $\angle ABD+ \angle AED=\beta+180-\beta=180$ $\implies$ $A,B,D,E-are$ $concyclic$

Define the intersection of DE with the circumcircle of ABC between arc BC and AC as B' and A', respectively. Let the circumcenter be O. Let $OC \cap DE=X$ Since DE is parallel to the tangent line, OC is perpendicular to DE, and bisects B'A'. This means triangles B'XC and A'XC are congruent so angles <XB'C and <XA'C are equal. Let the aforementioned angle have measure x. <XB'C and <CAA' both intercept arc A'C so <CAA'=x and by similar reasoning <B'BC=x. We also know that ABB'A' is cyclic so <BAA'+BB'A'=180. Let <BAC have measure A. This means <BAA'=A+x so BB'A'=180-A-x. This means <BDC'=180-(180-A-x)-x=A so <BDE=180-A which means ABDE is cyclic.