I draw a quadrilateral ABCD of which \( \mathrm{BC}=6 \mathrm{~cm}, \mathrm{AB}=4 \mathrm{~cm}, \mathrm{CD}=3 \mathrm{~cm}, \angle \mathrm{ABC}=60^{\circ} \), \( \angle \mathrm{BCD}=55^{\circ} \), I draw a triangle with equal area of that quadrilateral of which one side is along side AB and other side is along side BC .