The Total solution for NCERT class 6-12
In Fig. 9.30, D and E are two points on BC such that BD = DE = EC.
Show that ar (ABD) = ar (ADE) = ar (AEC).
Can you now answer the question that you have left in the ‘Introduction’of this chapter, whether the field of Budhia has been actually divided intothree parts of equal area?
Given,
BD = DE = EC
To prove,
ar (△ABD) = ar (△ADE) = ar (△AEC)
Proof,
In (△ABE), AD is median [since, BD = DE, given]
We know that, themedian of a triangle divides it into two parts of equal areas
, ar(△ABD) = ar(△AED) —(i)
Similarly,
In (△ADC), AE is median [since, DE = EC, given]
,ar(ADE) = ar(AEC)—(ii)
From the equation (i)and (ii), we get
ar(ABD) = ar(ADE) =ar(AEC)