A circle [MATH]k(O)[/MATH] with diameter [MATH]AB[/MATH] is given. Lines [MATH]PC[/MATH] and [MATH]PD[/MATH] touch k ([MATH]C, D \in k[/MATH]). [MATH]AC \cap BD = K[/MATH]. Show that [MATH]PK \bot AB[/MATH].
I have tried to calculate some angles if [MATH]\angle DCP = \angle PDC = \alpha[/MATH] but it seems useless at the end. I also see that [MATH]\angle ACB = \angle ADB = \frac {\overset{\mmlToken{mo}{⏜}}{AB\,} }{2} = 90 ^\circ[/MATH]. Also [MATH]\triangle ABC[/MATH] and [MATH]\triangle ABD[/MATH] share the side AB. I think they are congruent but I don't know how to show it. I will be very grateful if you help me!
I have tried to calculate some angles if [MATH]\angle DCP = \angle PDC = \alpha[/MATH] but it seems useless at the end. I also see that [MATH]\angle ACB = \angle ADB = \frac {\overset{\mmlToken{mo}{⏜}}{AB\,} }{2} = 90 ^\circ[/MATH]. Also [MATH]\triangle ABC[/MATH] and [MATH]\triangle ABD[/MATH] share the side AB. I think they are congruent but I don't know how to show it. I will be very grateful if you help me!
