Answer:
Given: AC, DF, and GI are parallel.
Solution:
In ΔJAB and ΔJDE
∠J is common.
∠JAB=∠JDE→→As, AB║DE, so corresponding angles are equal.
ΔJAB ~ ΔJDE→→[AA similarity]
Similarly, we can prove that, ΔJ C B and ΔJ FE by AA similarity.
As, we know when triangles are similar their sides are proportional.
[tex]\frac{JA}{JD}=\frac{JB}{JE}=\frac{AB}{DE}\\\\ \frac{JB}{JE}=\frac{JC}{JF}=\frac{BC}{EF}[/tex]
[tex]\frac{JA}{AB}=\frac{1}{JE}\\\\\frac{JC}{BC}=\frac{1}{JE}\\\\\frac{JA}{AB}=\frac{JC}{BC}\\\\= \frac{JA}{JC}=\frac{AB}{BC}[/tex]
Option 4: BC is correct choice.
