# Circle and triangle

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 $$KK_1=>M' \equiv M$$. • Welcome to Math.SE! Please read this post and the others there for information on writing a good question for this site. In particular, people will be more willing to help if you edit your question to include some motivation, and an explanation of your own attempts. – GNUSupporter 8964民主女神 地下教會 Apr 26 at 20:48
• @GNUSupporter8964民主女神地下教會, I'm still trying to think up sth but... – Nikol Dimitrova Apr 26 at 20:49
• @GNUSupporter8964民主女神地下教會, I've edited my post and shared everything that I succeeded to do. – Nikol Dimitrova Apr 26 at 21:06
• If you indeed think $B$ is the orthocenter, thrn what is it of $\Delta DPC$?(you also have tangents and radiuses) ;) – user665856 Apr 27 at 2:42
• I've edited the drawing and have shown some work. Now I think that's a correct post. – Nikol Dimitrova Apr 27 at 19:18

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 $$AK_1K$$.