In a quadrilateral ABCD,which is not a trapez | vedclass

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(C) Given: In quadrilateral $ABCD$,$\angle DAB = \angle ABC = 60^{\circ}$ and $\angle CAB = \angle CBD$.Construction: Extend $AD$ and $BC$ to meet at

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$AB = BC + AD$

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(C) Given: In quadrilateral $ABCD$,$\angle DAB = \angle ABC = 60^{\circ}$ and $\angle CAB = \angle CBD$.Construction: Extend $AD$ and $BC$ to meet at point $E$ such that $	riangle AEB$ is an equilateral triangle.Since $	riangle AEB$ is equilateral,$AB = BE = AE$.In $	riangle BED$ and $	riangle ABC$:$\angle E = 60^{\circ}$ (as $	riangle AEB$ is equilateral).$\angle ABC = 60^{\circ}$ (given).Thus,$\angle E = \angle ABC$.Also,$\angle DBE = \angle CAB$ (given).Therefore,by $AA$ similarity,$	riangle BED \sim 	riangle ABC$.From the similarity,we have the ratio of corresponding sides:$\frac{BE}{AB} = \frac{ED}{BC} = \frac{BD}{AC}$.Since $AB = BE$,the ratio $\frac{BE}{AB} = 1$.Therefore,$\frac{ED}{BC} = 1$,which implies $ED = BC$.From the construction,$AE = AD + ED$.Since $AE = AB$ and $ED = BC$,we substitute these into the equation:$AB = AD + BC$.

In a quadrilateral $ABCD$,which is not a trapezium,it is known that $\angle DAB = \angle ABC = 60^{\circ}$. Moreover,$\angle CAB = \angle CBD$. Then,

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