Suppose $ABCD$ is a cyclic quadrilateral and $P$ is the intersection of the lines determined by $AB$ and $CD$. Show that $PA·PB= PD·PC$
Could you help me please, I have no idea how to relate the property that he is cyclic, I have reviewed other posts of cyclic quadrilaterals and the truth that I have not managed to understand or how to attack this problem. Thank you.
This is a case of the power of a point theorem. You can prove it using similar triangles. By the inscribed angle theorem, $\angle BAC=\angle BDC$ and so $$\angle PAC=PDB.$$ Again by the inscribed angle theorem, $$\angle ABD=\angle ACD.$$ By AA similarity, this establishes that $$\triangle PCA \sim \triangle PBD.$$ By similarity ratios, $$\frac{PA}{PD}=\frac{PC}{PB},$$ which is what we wanted.