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In figure, D and E are two points on BC such that BD = DE = EC. Show that ar(ABD) = ar(ADE) = ar(AEC).

Let us draw AF, perpendicular to BC such that AF is the height of $$\triangle$$ABD, $$\triangle$$ADE and $$\triangle$$AEC

$$\therefore$$ area of triangle =( $$\frac{1}{2}$$) × base ×Height
$$\therefore$$ Area of $$\triangle$$ ABD= ($$\frac{1}{2}$$ ) × BD × AF
$$\therefore$$ Area of $$\triangle$$ ADE = ($$\frac{1}{2}$$ ) × DE × AF
$$\therefore$$ Area of $$\triangle$$ AEC = ($$\frac{1}{2}$$ ) × EC × AF
Since , BD= DE= EC (Given )
$$\therefore$$ Area of $$\triangle$$ ABD= Area of $$\triangle$$ ADE= Area of $$\triangle$$ AEC