# In the adjoining figure, DE ‖ BC. Prove that(i) ar(∆ACD) = ar(∆ABE),(ii) ar(∆OCE) = ar(∆OBD). Given

AB DC

To prove that : (i) area(∆ACD) = area(∆ABE)

(ii) area(∆OCE) = area(∆OBD)

(i)

Here in the given figure Consider BDE and ECD,

we find that they have same base DE and lie between two parallel lines BC and DE

According to the theorem: triangles on the same base and between same parallel lines have equal

areas. Area of BDE = Area of ECD

Now,

Area of ACD = Area of ECD + Area of ADE ---1

Area of ABE = Area of BDE + Area of ADE ---2 From 1 and 2

We can conclude that area(∆AOD) = area(∆BOC) (Since Area of ADE is common)

Hence proved

(ii)

Here in the given figure Consider BCD and BCE,

we find that they have same base BC and lie between two parallel lines BC and DE

According to the theorem : triangles on the same base and between same parallel lines have equal

areas. Area of BCD = Area of BCE

Now,

Area of OBD = Area of BCD - Area of BOC ---1

Area of OCE = Area of BCE - Area of BOC ---2 From 1 and 2

We can conclude that area(∆OCE) = area(∆OBD) (Since Area of BOC is common)

Hence proved

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