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Question

# In the given figure, line DE || line GF ray EG and ray FG are bisectors of $\angle$DEF and $\angle$DFM respectively. Prove that, (i) $\angle$DEG = $\frac{1}{2}\angle$EDF (ii) EF = FG.

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Solution

## (i) Given: DE || GF Now, $\angle$DEF = $\angle$GFM (Corresponding angles as DM is a transversal line) ⇒ 2$\angle$DEG = $\angle$DFG (Ray EG and ray FG are bisectors of $\angle$DEF and $\angle$DFM) ⇒ 2$\angle$DEG = $\angle$EDF (∵ $\angle$EDF = $\angle$DFG, alternate angles as DF is a transversal line) ⇒ $\angle$DEG = $\frac{1}{2}\angle$EDF (ii) Given: DE || GF $\angle$DEG = $\angle$EGF (Alternate angles as EG is a transversal line) ∴ $\angle$GEF = $\angle$EGF (∵ $\angle$DEG = $\angle$GEF) ∴ EF = FG (Sides opposite to equal angles)

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