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Question
In the given figure: AB//FD, AC//GE and BD = CE;
prove that:
- BG = DF
- CF = EG
Solution
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
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