inversion at D,and the constant point is the reflection of A' wrt ω'

2016 Oral Moscow Geometry Olympiad P5 wrote: From point $A$ to circle $\omega$ tangent $AD$ and arbitrary a secant intersecting a circle at points $B$ and $C$ (B lies between points $A$ and $C$). Prove that the circle passing through points $C$ and $D$ and touching the straight line $BD$, passes through a fixed point (other than $D$). After Inversion at $D$ with an arbitary radius the problem gets equivalent to this problem:- Inverted Problem wrote: A line contains points $A',D$ and a line $\|$ to $A'D$ be named as $\omega'$ and let $B'$ be an arbitary point on $\omega'$ and let $\odot(B'A'D)\cap\omega'=C'$. Let a line $\|$ to $B'D'$ through $C'$ intersect $A'D$ at $T'$. Then $C'T'$ passes through a fixed point other than $C'$ Let $T'C'\cap\odot(DA'B'C')=K.$. Then $$\angle A'DB'=\angle A'CB'=\angle A'T'K'=\angle B'C'K'\implies \text{Reflection of } A' \text{ over } \omega' \text{ lies on } C'T' \text{ which is the fixed point}$$ Hence, Inverting back we get that Midpoint of $DE$ is the fixed point, where $E$ is the Point of Tangency made by the second tangent from $A$ to $\omega$. $\qquad\blacksquare$

parmenides51 wrote: From point $A$ to circle $\omega$ tangent $AD$ and arbitrary a secant intersecting a circle at points $B$ and $C$ (B lies between points $A$ and $C$). Prove that the circle passing through points $C$ and $D$ and touching the straight line $BD$, passes through a fixed point (other than $D$). Let $E$ be the other point of tangency from $A$ to $\omega$.Thus $DBCE$ is harmonic.Let $M$ be the midpoint of $DE$.Thus \[\measuredangle{MCD}=\measuredangle{ECA}=\measuredangle{ECB}=\measuredangle{EDB}=\measuredangle{MDB}\].Thus $\odot{MDC}$ is tangent to $BD$ and since $M$ is fixed for a particular $A$ so is the required point.$\square$