Q. 585.0( 1 Vote )

# In figure, EF is a t

Answer :

Given: AB || CD and EF is a transversal cutting AB at G and CD at H. GM is the bisector of ∠ EGB and HL bisector of ∠ GHD. To prove: ∠ MGB = ∠ LHD. Proof: Since AB || CD and EF is a transversal. ∠ EGB = ∠ GHD (Corresponding angles are equal). --------------- (1) GM is the bisector of ∠ EGB ∴ ∠ EGM = ∠ MGB or ∠ MGB = ∠ EGB------------- (2) Similarly, HL is the bisector of ∠ GHD ∴ ∠ LHD = ∠ GHD ---------------(3) From (1)     Ð EGB = ∠ GHD  ∠ EGB = ∠ GHD ∴ ∠ MGB = ∠ LHD [From (2) and (3)] GM and HL are two rays cut by a transversal EF and ∠ MGB = ∠ LHD. ∴ GM || HL If a transversal intersects two lines in such a way a pair of corresponding angles are equal then the two lines are parallel.

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