Q. 6

# In figure, if EF

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

AGD = AIE |Right Angle

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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