In ΔABC, if DE divides AB and AC in the same ratio, then which of the following options is true?
DE and BC are parallel
Construction: Draw DE' parallel to BC as shown in the figure above.
Since, DE' II BC,
ADDB=AE′E′C ... (i) [by Basic Proportionality Theorem]
But we are given that AD divides AB and AC in the same ratio then
ADDB=AEEC ...(ii)
From (i) and (ii), we get
AE′E′C=AEEC
Adding 1 to both sides further, we get
AE′E′C+1=AEEC+1
ACE′C=ACEC
E′C=EC
This is possible only if E' and E coincide. (Since E' and E lie on the same line)
This implies DE' = DE
Hence, DE ∥ BC.