a) Ta có: \(AP\cdot AQ=AB^2\)
\(\Leftrightarrow AP\cdot AQ=AB\cdot AC\)
hay \(\dfrac{AP}{AC}=\dfrac{AB}{AQ}\)
Xét ΔAPB và ΔACQ có
\(\dfrac{AP}{AC}=\dfrac{AB}{AQ}\)(cmt)
\(\widehat{PAB}=\widehat{CAQ}\)\(\left(=\widehat{PAx}\right)\)
Do đó: ΔAPB\(\sim\)ΔACQ(c-g-c)