There was a Finnish matriculation examination there was the following question:
Consider a circle and a point $P$ outside the circle. From the point $P$ draw two lines such that each of the line intersects the circle twice, at the points $A,B,C,D$ where $P, A, B$ are collinear as well as $P, C, D$ are collinear. (The original link is http://www.mafyvalmennus.fi/images/uploads/pmyo_s14.pdf problem 14, but as of 2023 this has gone dead and there is no Internet Archive image.)
a) Prove that the triangles $PCB$ and $PAD$ are similar.
b) Prove that $PA\cdot PB=PC\cdot PD$.
c) Consider the special case $A=B$. Prove that in this case $(PA)^2=PC\cdot PD$.
d) Prove using a)-c) Pythagorean theorem.
Is this solution I found on the net valid for c)?
Triangles $PAD$ and $PCA$ has still one common angle $APD$. By the inscribed angle theorem, the inscribed angles $PAC$ and $PDA$ corresponding the same arc are equal so triangles $PAD$ and $PCA$ has two equal angles, and they are similar. Therefore, $PA/PD=PC/PA$ so $(PA)^2=PC\cdot PD$.
What makes me wonder is that can we use the inscribed angle theorem as $P$ is outside the circle. Is this proof valid?
Inscribed angle theorem is for angles which are in the interior of a circle, however it is not important for this problem, since we can consider this case, which $PA$ is tangent to the circle, as a limit case of interior angle theorem. For understanding the problem better, consider the previous case, in which $A$ and $B$ do not coincide with each other. In that case we have:
$\measuredangle{PAC}=\pi-\measuredangle{BAC}=\pi-\frac{1}{2}\widetilde{BDC}=\frac{2\pi-\widetilde{BDC}}{2}=\frac{\widetilde{BAC}}{2}$
in which by "$\thicksim$" we mean "arc".
Now we take the limit of the above relation as $B\rightarrow A$.
$\lim_{B\rightarrow A} \measuredangle{PAC}=\lim_{B\rightarrow A} \frac{\widetilde{BAC}}{2}=\frac{\widetilde{AC}}{2}$.
As $\measuredangle{ADP}=\frac{\widetilde{AC}}{2}$, we can claim that in this case that $\overline{PA}$ is tangent to the circle, $\measuredangle{PAC}=\measuredangle{ADP}$, and as $\measuredangle{ADP}$ is common between both triangles $PAC$ and $PAD$, hence those triangles are similar.