If two triangles are on a common base and have equal areas, then they will lie between the same parallel lines. Let's draw the given triangle ABC. Since ΔEBC and ΔDBC are lying on a common base BC and they also have equal areas, therefore, ΔEBC and ΔDBC will lie between the same parallel lines. Hence, DE || BC proved.
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Answers & Comments
Step-by-step explanation:
Given that D and E are points on sides AB and AC respectively of △ABC such that ar.(△DBC)=ar.(△EBC)
To Prove: DE∥BC.
Construction: Join BE and DC.
Proof:
△DBC and △EBC are on base BC.
ar.(△DBC)=ar.(△EBC) (Given)
Using the fact that two triangles on the same base (or equal bases) and between the same parallel lines are equal in area, we can say that:
BC∥DE
Hence, proved.
Verified answer
Answer:
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If two triangles are on a common base and have equal areas, then they will lie between the same parallel lines. Let's draw the given triangle ABC. Since ΔEBC and ΔDBC are lying on a common base BC and they also have equal areas, therefore, ΔEBC and ΔDBC will lie between the same parallel lines. Hence, DE || BC proved.
[tex]\huge\color{purple}{\colorbox{orange}{\colorbox{white}{ÃÑẞWËR BY PrēēT}}}[/tex]
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