Q6 of 56 Page 119

In figure, if EF || DC || AB then prove that


Let us drop a perpendicular AG and BH to CD cutting EG at I and J and CD.


In ΔADG & ΔAEI,



∠AGD = ∠AIE |Right Angle


∠AEI = ∠ADG |corr. ∠s


ΔADG~ΔAEI by AA Similarity Rule



In ΔBJF & ΔBHC,



∠BJF = ∠BHC |Right angle


∠BFJ = ∠BCH |corr. ∠s


ΔBJF~ΔBHC by AA Similarity Rule



In rectangle ABHG & ABJI,


AI = BJ …(a) |opposite sides of rectangle are equal


AG = BH …(b) |opp. sides of rectangle


From eqn. (b) – (a)


AG – AI = BH – BJ


⇒ GI = HJ



…(2)


From (1) & (2),



Hence, proved.


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