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Question

# Prove that parallelogram on the same base and between the same parallels are equal in area.

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Solution

## Given: The parallelograms ABCD and ABEF be on the same base AB and between same parallels AB and FC. ​To prove: ar(ABCD) = ar(ABEF) Proof: $\mathrm{As},\mathrm{CD}=\mathrm{AB}\left(\mathrm{Opposite}\mathrm{sides}\mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}\right)\phantom{\rule{0ex}{0ex}}\mathrm{And},\mathrm{FE}=\mathrm{AB}\left(\mathrm{Opposite}\mathrm{sides}\mathrm{of}\mathrm{parallelogram}\mathrm{ABEF}\right)\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{FE}=\mathrm{CD}\phantom{\rule{0ex}{0ex}}⇒\mathrm{FE}+\mathrm{ED}=\mathrm{CD}+\mathrm{ED}\phantom{\rule{0ex}{0ex}}⇒\mathrm{FD}=\mathrm{CE}.....\left(\mathrm{i}\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\mathrm{In}∆\mathrm{ADF}\mathrm{and}∆\mathrm{BCE},\phantom{\rule{0ex}{0ex}}\mathrm{AD}=\mathrm{BC}\left(\mathrm{Opposite}\mathrm{sides}\mathrm{of}\mathrm{parallelogram}\mathrm{ABCD}\right)\phantom{\rule{0ex}{0ex}}\mathrm{AF}=\mathrm{BE}\left(\mathrm{Opposite}\mathrm{sides}\mathrm{of}\mathrm{parallelogram}\mathrm{ABEF}\right)\phantom{\rule{0ex}{0ex}}\mathrm{DF}=\mathrm{CE}\left[\mathrm{From}\left(\mathrm{i}\right)\right]\phantom{\rule{0ex}{0ex}}\mathrm{So},\mathrm{by}\mathrm{SSS}\mathrm{congruency}\phantom{\rule{0ex}{0ex}}∆\mathrm{ADF}\cong \mathrm{BCE}\phantom{\rule{0ex}{0ex}}⇒\mathrm{ar}\left(∆\mathrm{ADF}\right)=\mathrm{ar}\left(∆\mathrm{BCE}\right)\left(\mathrm{Congruent}\mathrm{triangles}\mathrm{are}\mathrm{equal}\mathrm{in}\mathrm{area}\right)\phantom{\rule{0ex}{0ex}}⇒\mathrm{ar}\left(\mathrm{ABCF}\right)-\mathrm{ar}\left(∆\mathrm{ADF}\right)=\mathrm{ar}\left(\mathrm{ABCF}\right)-\mathrm{ar}\left(∆\mathrm{BCE}\right)\phantom{\rule{0ex}{0ex}}\therefore \mathrm{ar}\left(\mathrm{ABCD}\right)=\mathrm{ar}\left(\mathrm{ABEF}\right)$

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