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Q.3. (b) in the figure given above, DE ‖ BC. Prove that:

(i) area of ∆ ACD = area of ∆ ABE

(ii) area of ∆ OBD = area of ∆ OCE

Please find below the solution to the asked query:

We know : Area of triangle = $\frac{1}{2}$ $\times $ Base $\times $ Height

Let , Height between parallel lines ( DE | | BC ) =

*h*

So,

Area of $\u2206$ BDE = $\frac{1}{2}$ $\times $ DE $\times $

*h*--- ( 1 ) ( Here base = DE and Height =

*h*as that is lies between DE | | BC )

And

Area of $\u2206$ CDE = $\frac{1}{2}$ $\times $ DE $\times $

*h*--- ( 2 ) ( Here base = DE and Height =

*h*as that is lies between DE | | BC )

From equation 1 and 2 we get :

Area of $\u2206$ CDE = Area of $\u2206$ BDE --- ( 3 )

i ) Now we add " Area of $\u2206$ ADE " on both hand side of equation 3 and get :

Area of $\u2206$ CDE + Area of $\u2206$ ADE = Area of $\u2206$ BDE + Area of $\u2206$ ADE

**Area of $\u2206$ ACD = Area of $\u2206$ ABE ( Hence proved )**

ii ) Now we subtract " Area of $\u2206$ ODE " on both hand side of equation 3 and get :

Area of $\u2206$ CDE - Area of $\u2206$ ODE = Area of $\u2206$ BDE - Area of $\u2206$ ODE

**Area of $\u2206$ OCE = Area of $\u2206$ OBD ( Hence proved )**

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