Extend BP and AD to meet at a point Q outside the quadrilateral Let angle BAD= angle BCD= x angle ABP= angle PBC= y So, angle AQB= 180°-(x+y) angle BPC= 180°-(x+y) angle BPC= angle DPQ= Vertically opposite angles angle DPQ=180°-(x+y) So, DP=DQ DP=CD/2=DQ AD=CD/3 So, AQ=5CD/6 In∆ABQ, sin(x+y)= (AB sin y)/AQ In∆PBC, sin(x+y)= (BC sin y)/CP AB/ AQ= BC/CP AB/BC= AQ/CP = AQ/ (CD/2 )= 2(AQ/CD)= 2*(5/6)= 5/3 AB:BC=5:3 @Krishijivi