Question

Question no. 6 ...............please

Answer 1

In given figure join AC and BD

In traingle ACP and DBP

angle APC =angle DPB...(comman angle)

angle BAC =angle BDC...(angle inscribed in same arc)

∆ACP~∆DBP

AP/DP=PC/BP...(c.s.s.t)

PA×PB=PD×PC

Hope it helps.........