C-14-Q025mediumsingle_mcqCorollary 1 of Theorem 28 gives —aABAD=ACAE\dfrac{AB}{AD} = \dfrac{AC}{AE}ADAB=AEACbADAB=AEEC\dfrac{AD}{AB} = \dfrac{AE}{EC}ABAD=ECcABDB=AEAC\dfrac{AB}{DB} = \dfrac{AE}{AC}DBAB=ACdABAD=ACCE\dfrac{AB}{AD} = \dfrac{AC}{CE}ADAB=CE