A circle $k(O)$ with diameter $AB$ is given. Lines $PC$ and $PD$ touch $k$ ($C, D \in k)$. $AC$ $\cap$ $BD =K$. Show that $PK \bot AB$.
This is what I've done for two days. I will be very grateful if someone can tell me is this enough because I'm not entirely sure.
$$\angle ACB = \angle ADB = \frac {\overset{\mmlToken{mo}{⏜}}{AB\,}}{2} = 90 ^\circ$$
Let's look in $\triangle AK_1K$: $$B\in K_1C; \angle K_1CA = \angle BCA = 90 ^\circ$$ $$B \in KD; \angle KDA = \angle BDA = 90 ^\circ$$ $=>B$ $-$ orthocenter $=> AB \bot KK_1$
$t_c$ $\cap$ $KK_1 = M$
$\angle ABC = \angle CKK_1$ (angles with perpendicular sides); $\angle KCM = \angle ECA = \frac {\overset{\mmlToken{mo}{⏜}}{AC\,}}{2} = \angle ABC$
$=> \angle CKK_1 = \angle KCM => \triangle CMK$ is isosceles $=> CM = KM$
$\angle CK_1M = \angle CAB$ (angles with perpendicular sides); $\angle K_1CM = \angle BCM = \frac {\overset{\mmlToken{mo}{⏜}}{BC\,}}{2} = \angle BAC$ $=> \angle CK_1M = \angle K_1CM => \triangle CMK_1$ is isosceles $=> CM = K_1M$
$=> KM = K_1M => M$ is midpoint of $KK_1$
Similarly: $t_d$ $\cap$ $KK_1=M'$, $M'$ is midpoint of $KK_1=>M' \equiv M$.
Note that as $PD,PC $ are tangents, hence, $PDOC$ is cyclic. So, $\angle DPC=180-2x$, where $x= \angle DAC$. Also, $\angle DBC=180-x$, and notice that proving your problem is equivalent to proving $B$ is the orthocenter of $\Delta AK_1K$(stronger even). And so, if it must be true, then $\Delta DPC$ must be the orthic triangle, and hence also, $B $ must be its incenter. Now note $\angle DPC=180-2x$, $\angle DBC=180-x$. And $90+0.5(\angle DPC)=90+0.5(180-2x)=90+90-x=(180-x)=\angle DBC$. So we are closer and just need to prove one angle bisection exists. But $PD=PC$ , and $DO=OC$, hence if we let $PO \cap CD = L$, then $PL$ bisects $\angle DPC$(by angle bisector theorem). So it is enough to force that $B$ is indeed the incenter of $\Delta DPC$. (As in the line in the incenter exists, only one point satisfies $\theta = 90+ 0.5(\alpha)$, and we have shown this condition and also proved $B$ lies in the line in which the incenter lies) So, it is the orthocenter of its extriangle $AK_1K$.