Question

In the given figure: AB//FD, AC//GE and BD = CE;

prove that: 

1. BG = DF     
2. CF = EG   

 

Answer 1

In the given figure AB || FD,

⇒ ∠ABC =∠FDC

Also AC || GE,

⇒ ∠ACB = ∠GEB

Consider the two triangles ΔGBE and ΔFDC

∠B = ∠D             ...(Corresponding angle)

∠C = ∠E             ...(Corresponding angle)

Also given that

BD = CE

⇒ BD + DE = CE + DE

⇒ BE = DC

∴ By Angle-Side-Side-Angle criterion of congruence

ΔGBE ≅ ΔFDC

∴ `"GB"/"FD" = "BE"/"DC" = "GE"/"FC"`

But BE = DC

⇒ `"BE"/"DC" = "BE"/"BE"` = 1

∴ `"GB"/"FD" = "BE"/"DC"` = 1

⇒ GB = FD

∴ `"GE"/"FC" = "BE"/"DC"` = 1

⇒ GE = FC