I draw a quadrilateral $A B C D$ of which $B C = 6 c m, A B = 4 c m, C D = 3 c m, ∠ A B C = 60^{\circ}$ , $∠ B C D = 55^{\circ}$ , I draw a triangle with equal area of that quadrilateral of which one side is along side $A B$ and other side is along side $B C$ .

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Solution

A. Draw the given quadrilateral $A B C D$ .
B. Draw the diagonal $A C$ of quadrilateral $A B C D$ .
C. Draw a parallel line through point $D$ to diagonal $A C$ of quadrilateral $A B C D$ which intersects at E produced BC.
D. $D E ∥ A C, △ A B E$ is the required triangle.
Proof:
$△ A C E = △ A D C$ (on same base $A C$ and between same parallels $A C$ and $D E$ )
$∴ △ A C E = △ A D C$
$△ A B C + △ A D C = △ A C E + △ A B C$ (adding area of $△ A B C$ on both sides)
$∴$ quadrilateral $A B C D = △ A B E$

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