Mathematics
In the given figure, D is the mid-point of BC and E is any point on AD. Prove that :
(i) ar (△EBD) = ar (△EDC)
(ii) ar (△ABE) = ar (△ACE)
Theorems on Area
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Answer
(i) Given,
BD = DC
Thus, ED is the median of triangle EBC.
We know that,
Median ED divides a triangle EBC into two triangles EBD and ECD of equal area.
∴ ar (△EBD) = ar (△EDC)
Hence, proved that ar (△EBD) = ar (△EDC).
(ii) Given,
BD = DC
Thus, AD is the median of triangle ABC.
Median AD divides a triangle ABC into two triangles ABD and ADC of equal area.
ar (△ABD) = ar (△ADC)
ar (△ABE) + ar (△EDB) = ar (△ACE) + ar (△EDC)
ar (△ABE) + ar (△EDB) = ar (△ACE) + ar (△EDB) [ar (△EBD) = ar (△EDC)]
ar (△ABE) + ar (△EDB) - ar (△EDB) = ar (△ACE)
ar (△ABE) = ar (△ACE).
Hence, proved that ar (△ABE) = ar (△ACE).
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