# Suppose ABCD is a cyclic quadrilateral and P is the intersection of the lines determined by AB and CD. Show that PA cdot PB=PD cdot PC.

Suppose ABCD is a cyclic quadrilateral and P is the intersection of the lines determined by AB and CD. Show that $PA·PB=PD·PC$.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

dkmc4175fl
Step 1
This is a case of the power of a point theorem. By the inscribed angle theorem, $\mathrm{\angle }BAC=\mathrm{\angle }BDC$ and so $\mathrm{\angle }PAC=PDB.$.
Again by the inscribed angle theorem, $\mathrm{\angle }ABD=\mathrm{\angle }ACD.$.
Step 2
By AA similarity, this establishes that $\mathrm{△}PCA\sim \mathrm{△}PBD.$.
By similarity ratios, $\frac{PA}{PD}=\frac{PC}{PB},$ which is what we wanted.