Q : 11     In Fig. \small 9.27, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that

(i)  \small ar(ACB)=ar(ACF)
(ii)  \small ar(AEDF)=ar(ABCDE)

        

Answers (1)
M manish


We have a pentagon ABCDE in which BF || AC and CD  is produced to F.

(i) Since \DeltaACB and \DeltaACF are on the same base AC and between same parallels AC and FB.
\therefore  ar(\DeltaACB) = ar (\DeltaACF)..................(i)

(ii) Adding the ar (AEDC) on both sides in  equation (i),  we get

  ar(\DeltaACB) + ar(AEDC) =  ar (\DeltaACF) + ar(AEDC)
 \therefore \small ar(AEDF)=ar(ABCDE)

Hence proved.

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