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In Figure, A B C D E\ is a pentagon. A ...

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)`

Text Solution

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i) `triangle ACB` and `triangleACF` are lying on the same base `AC` and are existing between the same parallels `AC` and `BF`
According to Theorem 9.2: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.
Therefore, `ar (triangleACB) = ar (triangleACF)`
ii) It can be observed that`ar (triangleACB) = ar (triangleACF)`
Adding area (ACDE) on both the sides.
Area `(triangleACB)` + Area (ACDE) = `Area (triangleACF)` + Area (ACDE)
Area (ABCDE) = Area (AEDF)
Hence proved.
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