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    				| 1. | In Figure, `A B C D E `is a pentagon. A line through `B`parallel to`A C`meets `D C`produced at`Fdot`Show that:`a r ( A C B)=a r ( A C F)``a r (A E D F)=a r (A B C D E)` | 
| Answer» Join AF. `triangleACB and triangleACF` have the same base AC and lie between the same parallels AC and BF. `therefore ar(triangleACB)=ar(triangleACF)`. And so, `ar(triangleACB)+ar(triangleACDF)=ar(triangleACF)+ar(triangleACDE)` `rArr ar(ABCDE)=ar(AEDF)`. | |