MN is a diameter, which means $MA \perp NA,MZ \perp NZ$ so $MAZN$ can not be a isosceles trapezium, it should be a square.

gnoka wrote: MN is a diameter, which means $MA \perp NA,MZ \perp NZ$ so $MAZN$ can not be a isosceles trapezium, it should be a square. If $MAZN$ is a square $\Rightarrow \angle MAZ = 90^{\circ}$ But, as you said, $MA \perp NA \Rightarrow \angle MAN = 90^{\circ}.$ $\angle MAZ = \angle MAN + \angle NAZ = 90^{\circ} + \angle NAZ$ (Contradiction since $\angle NAZ \neq 0$) Not my sketch:

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Since $DF$ is a dimeter, we have $EF \perp DE$, so $ \angle PED = 90^o = \angle AEC $. On the other hand, $\angle ACE = \angle EAB = \angle EAZ = \angle EDP$ since $AEDZ$ is cyclic. Therefore, triangles $AEC$ and $PDE$ are spirally similar. So, we have $\angle EPD = \angle EAC = \angle ABE$, implying both $PEAC$ and $PEBZ$ are cyclic. From $PEAC$ cyclic we have $\angle APC = \angle AEC = 90^o$. This means that $B$ is the circumcentre of triangle $APZ$. ($\angle APZ = 90^o$ and $B$ is the midpoint of hypotenuse $AZ$) So, $BP=BA$. Then we have $\angle TAZ = \angle PAB = \angle APB = 180^o - \angle APC - \angle BPZ = 180^o - 90^o - \angle BPZ = 180^o - \angle AEB - \angle BEZ = 180^o - \angle AEZ = \angle AZT$, the last equality is true since $TZ$ is a tangent. Therefore, $TA = TZ$, so $TA$ and $TZ$ are both tangent to the circle $(c)$. Now, by La Hire's Theorem, if $TM$ intersects $(c)$ at $X \neq M$, then $KM$ and $KX$ are both tangents to $(c)$. So $KMOX$ is a kite, and therefore $OK \perp XM$, which implies $OK \perp TM$. (proven) Not sure why $MAZN$ has to be an isosceles trapezium though.

Obviously if $ TA $ is tangent to $\odot O$ then $ KBOM $ is a rectangle, so $\frac{KM}{MO}=\frac{OB}{OA}=\frac{OA}{OT}=\frac{OM}{OT}$ which implies that $\triangle KMO \sim \triangle MOT$, thus $\angle TMO + \angle KOM = \angle OKM + \angle KOM = 90^ {\circ} \Rightarrow OK \perp TM$ So we just have to prove that $TA$, or $PA$ is tangent to $\odot O$ It is easy to see that $\angle DEF = 90^{\circ}$, so $\angle DEP = 90^{\circ}$ Since $\angle ACP + \angle AEP = 90^{\circ} - \angle AZD +270^{\circ} - \angle AED = 90^{\circ} - \angle AZD + 90^{\circ} + \angle AZD = 180^{\circ}$, we have that $A, C, P, E$ are concylic So $\angle PAE = \angle PCE = \angle ZCE$ Take $G$ on line BC such that $ EB = BG $, then $AEZG$ is a parallelogram as $ B $ is the midpoint of $ AZ $ So $\angle ZGC = \angle GEA = 90^{\circ} = \angle ZAC$, which implies that $A, C, Z, G$ are concylic So $\angle ZCE = \angle ZCG = \angle ZAG$, and by $EZ \parallel AG$ we have $\angle ZAG = \angle AZE$ Finally we get $\angle PAE = \angle AZE$, which by the tangent-chord theorem is equivalent to $ PA $ is tangent to $\odot O$ $QED$

Why so many spurious points grah Delete $K$ and $T$ almost immediately by trivial angle chases to show that it STP $AP$ tangent to $(MANZ)$. Now redefine $P=(CAE)\cap(BZE)\neq E$; by an angle chase $\measuredangle CPE=\measuredangle CAE=\measuredangle ABC=\measuredangle ZBE=\measuredangle ZPE$, so we only need to show that $P\in\overline{EF}$. Angle chasing again forces us to first prove $AF=AZ$. However it’s clear if $A’$ is the point diametrically opposite $A$ on $(MANZ)$ then $AA’ZC$ parallelogram, so $\overline{FZ}\perp\overline{DZ}\parallel\overline{AA’}$ and we are done. From here we claim we win because $\measuredangle PAE=\measuredangle PCE=\measuredangle AA’E$ which is enough.