We extend $TM,TN$ to meet $\Omega_2$ at $P,Q$ respectively. Then $PQ\parallel MN$. Sine rule on $\triangle AMK$ yields $$MK = \frac{AM \sin{\angle MAK}}{\sin{\angle AKM}}$$Sine rule on $\triangle KND$ yields $$\frac{1}{KN}= \frac{\sin{\angle DKN}}{DN \sin{\angle KDN}}$$Sine rule on $\triangle MKB$ yields $$MK = \frac{MB \sin{\angle MBK}}{\sin{\angle MKB}}$$Sine rule on $\triangle NKC$ yields $$\frac{1}{NK} = \frac{\sin{\angle NKC}}{NC\sin{\angle NCK}}$$Multiplying the four equations, we get $$(\frac{MK}{KN})^2= \frac{AM \cdot BM}{CN \cdot DN}\cdot \frac{\sin{\angle MAK}}{\sin{\angle KDN}} \cdot \frac{\sin{\angle DKN}}{\sin{\angle MKB}} \cdot \frac{\sin{\angle MBK}}{\sin{\angle NCK}} \cdot \frac{\sin{\angle NKC}}{\sin{\angle AKM}}$$But the four pairs of angles above are all equal, so $$(\frac{MK}{KN})^2= \frac{AM \cdot BM}{CN \cdot DN}=\frac{TM\cdot PM}{TN\cdot QN}=(\frac{TM}{TN})^2$$since $PQ\parallel MN$. And so $$\frac{MK}{KN}=\frac{TM}{TN}$$and we are done.

Let $TM$ and $TN$ meet $\Omega_2$ at $S$ and $R$ and Let $\frac{TM}{TN} = x$. By Stewart Theorem we have $KM^2 = \frac{TM.SK^2 + SM.KT^2}{TS} - TM.MS$ and $KN^2 = \frac{TN.KR^2 + RN.TK^2}{TR} - TN.NR$ and as $T$ is touch point then $SR || MN$ so now using $\frac{TM}{TN} = x$ on $\frac{KM^2}{KN^2}$ we have $\frac{KM^2}{KN^2} = x^2 \implies \frac{KM}{KN} = x = \frac{TM}{TN} \implies TK$ is angle bisector.

Let $MN$ meets $\Omega_1$ at $U,W$ and $X=TT\cap UW$ $K=AC\cap BD \cap UW$ from $DIT$ on quadrilateral $ABCD $ and conic $\Omega_1$ there is an involution $(U,W)(M,N)(K,K)$ but we can see that inversion centered $X$ with radius $XT$ swaps $(U,W)(M,N)$ so $XT=XK$ which gives $TK$ is angle bisector of $\angle MTN$ $\blacksquare.$