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Question
The vertex A of △ ABC is joined to a point D on the side BC.The midpoint of AD is E.Prove that ar(△BEC)=12ar(△ ABC).
The vertex A of △ ABC is joined to a point D on the side BC.The midpoint of AD is E.Prove that ar(△BEC)=12ar(△ ABC).
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Solution
Given: D is the midpoint of BC and E is the midpoint of AD.
To prove: ar(ΔBEC)=
ar(ΔABC)
Proof:
Since E is the midpoint of AD, BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED ) = 
ar(∆ABD) ...(i)
Also, ar(∆CDE ) =
ar(∆ADC) ...(ii)
From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE) = 
⨯ ar(∆ABD) + 
⨯ ar(∆ADC)
⇒ ar(∆BEC ) = 
⨯ [ar(∆ABD) + ar(∆ADC)]
⇒ ar(∆BEC ) = 
⨯ ar(∆ABC)
To prove: ar(ΔBEC)=
ar(ΔABC)Proof:
Since E is the midpoint of AD, BE is the median of ∆ABD.
We know that a median of a triangle divides it into two triangles of equal areas.
i.e., ar(∆BED ) =
ar(∆ABD) ...(i)Also, ar(∆CDE ) =
ar(∆ADC) ...(ii)From (i) and (ii), we have:
ar(∆BED) + ar(∆CDE) =
⨯ ar(∆ABD) + ⨯ ar(∆ADC) ⇒ ar(∆BEC ) =
⨯ [ar(∆ABD) + ar(∆ADC)] ⇒ ar(∆BEC ) =
⨯ ar(∆ABC)Suggest Corrections
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