Abcd is a trapezium in which ab||dc and e, f are points on ad and bc respectively such that ef||ab or dc. Prove that ae/ed = bf/fc
Answers
Answer:
Let us join AC to intersect EF at G .
AB // DC and EF // AB ( given )
=> EF // DC
[ Lines parallel to the same line are
parallel to each other. ]
In ∆ADC , EG // DC
AE/ED = AG/GC ---- ( 1 )
[ By Basic Proportionality theorem ]
Similarly ,
In ∆CAB , GF // AB
CG/GA = CF/FB
[ By basic Proportionality theorem ]
AG/GC = BF/FC ---( 2 )
from ( 1 ) and ( 2 ) ,
AE/ED = BF/FC
=> AE/BF = ED/FC
Hence proved.
I hope this helps you.
Hi mate! here's your answer,
ABCD is trapizium in whichAB||DC and P,Q are mid points of AD, BC respectively.
Join CP and produce it to meet BA produced at R.
In ∆ PDC and ∆PAG,
PD =PA (P is mid point of AD)
CPD = RPC (Vertically opposite angles)
PCD = PRA (alternate angles)
(DC || AB, DC || RB and CR is transversal)
.•.∆PDC = ∆PAR
CD = RA and PR = PR
In ∆ CRB,
P is midpoint of CR( PC=PR proved )
Q is a midpoint of BC (given)
Therefore, By midpoint theoerm PQ || AB and PQ = (1/2) RB.
But RB = RA+AB = CD+AB
Hence, PQ || AB and PQ = (1/2) ( AB+CD).
Hope this will help you.
Mark as brainlist if useful for you.
