easy problem . First of all I think that there's a mistake , $P,Q,R$ must be the midpoints of $BC,CA,AB$ respectively... Now because: $PRAQ$ is a parallelogram and because $\angle ETP=\angle ESP=90$ so $ETPS$ is cyclic and $\angle TES=180-\angle TPS=180-\angle RAQ =180-\angle BAC = \angle T^{'}AS^{'}$ and because $AT^{'}E^{'}S^{'}$ is cyclic too, so the quadrilaterals $AT^{'}E^{'}S^{'}$ and $ETPS$ are similar. Now assume that $EP\cap TS=D$ and $ST\cap S^{'}T^{'}=G$ and $S^{'}T^{'}\cap AE^{'}=J$. because $AT^{'}E^{'}S^{'}$ and $ETPS$ are similar so we have $\angle ADG=\angle EDT=\angle T^{'}JA=180-\angle AJG$ , so $AJGD$ is cyclic too so because $FF^{'}$ is a diameter of the circumcircle of $ABC$ we have $\angle JAD=\angle F^{'}AF=90$ , and because $AJGD$ is cyclic we have : $\angle JGD=90$ and that means $SG$ or $ST$ is perpendicular to $S^{'}T^{'}$.

sinajackson wrote: Easy problem : $PRAQ$ is a parallelogram and because $\angle ETP=\angle ESP=90$ so $ETPS$ is cyclic and $\angle TES=\ldots =\angle T^{'}AS^{'}$ and because $AT^{'}E^{'}S^{'}$ is cyclic too, so the quadrilaterals $AT^{'}E^{'}S^{'}$ and $ETPS$ are similar. And yet, why ... are similarly ?

Well because $\angle AT^{'}E^{'}=\angle ETP,\angle AS^{'}E^{'}=\angle ESP, \angle T^{'}AS^{'}=\angle TES$ and $\angle T^{'}E^{'}S^{'}=\angle TPS$ , .... is there a problem?

Sorry, I think was sleeping then. It is all O.K.

To Sinajackson : I corrected the problem, but it's different with what you said.

And yet, I am right ! $A=A'\ ,\ B=B'\ ,\ C=C'\ ,\ D=D'\not\Longrightarrow ABCD\sim A'B'C'D'$, i.e. exist at least two quadrilaterals with the same angles and which yet aren't similarly.

Yes, SinaJackson's solution is not right.

Hello Mr Hatami and Virgil Nicula.... I think my solution is correct, I agree that we can find two quadrileterals that have the same angles and are not similar.... but here there's a difference, we have two right angles in these quadrilaterals... I think my solution is correct.... ( sorry for having much confidence )

Example for Sinajackson. Quote: Let $ABCD$ be a fixed convex quadrilateral inscribed in the circle $w(O)$ with the diameter $[AC]$, i.e. $B=D=90^{\circ}$ Let $\{X,Y\}\subset w$ be two mobile points so that $XY=BD$, the line $AC$ does not separate the points $B$, $X$ and the line $XY$ separates the points $A$, $O$. Then the convex cyclic quadrilaterals $ABCD$ and $AXCY$, where $B=X=D=Y=90^{\circ}$, have the same values of the its (corresponding to the writing) angles, i.e. and $\widehat{BAD}\equiv\widehat{XAY}$, $\widehat{BCD}\equiv\widehat{XCY}$ and yet they never aren't similarly, i.e. $ABCD\not\sim AXCY\ .$

But they have all their edges perpendicular ...

sorry , but I still think my solution is ok, and I think it is completely obvious that we can use the similarity of the two cyclic quadrilaterals in my solution.... your example is a completely special one that will not occur every time......

Are agree you with the my example ? I didn't say that the your proof is wrong, but that you used a false assertion.

Hm...what do you need similar quadrilaterals for? My solution: We have, since $TESP$ is cyclic \[\angle T'AS'=180-\angle CAB=180-\angle TPS=\angle TES.\] Also, since $E'A\perp AP$ and $T'A\perp T'E'$, we have \[\angle EST=\angle EPT=\angle PAQ=\angle AE'T'=\angle AS'T'.\] Hence the triangles $AT'S'$ and $ETS$ are similar (with the same orientation). So there exists a spiral similarity taking $ETS$ to $AT'S'$. However the corresponding sides $ES$ and $AS'$, as well as $ET$ and $AT'$ are perpendicular, so the third pair of sides, $TS$ and $T'S'$ must also be perpendicular.

After All this is an Iranian Problem : sinajackson wrote: Well because $\angle AT^{'}E^{'}=\angle ETP,\angle AS^{'}E^{'}=\angle ESP, \angle T^{'}AS^{'}=\angle TES$ and $\angle T^{'}E^{'}S^{'}=\angle TPS$ , .... is there a problem? Yes there is. Sina , Livi (Virgil Nicula) gave you a counterexample. Here is another one : Every rectangle has $4$ right angles , Do you think they are all Similar? I dont think so. If you still dont believe me , Prove your claime rigorously. Yimin ge ,I solved the problem like you. I wonder how they created this problem ?

Can somebody please post a diagram for this problem?

I think the difficulty of this problem is that we have to consider various configurations. For a specific case, the problem is easy.