The correct option is A True
InABD,CDisbisectedbyABatP.Therefore,APisthemedianofABD.Weknowamediandividestheareaofatriangleinhalf.∴Area(ABP)=Area(ADP)−−−(1)InBCD,BDisbisectedbyACatP.Therefore,APisthemedianofABD.Weknowamediandividestheareaofatriangleinhalf.∴Area(BCP)=Area(DCP)−−−(2)Addingequations1&2,wegetArea(ABP)+Area(BCP)=Area(ADP)+Area(DCP)⇒Area(ABC)=Area(ADC)