Suppose ABCD (AB parallel CD) is a trapezium | vedclass

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(C) Given that $AC$ is the bisector of $\angle DAB$,we have $\angle DAC = \angle CAB$.Since $DC \parallel AB$,the alternate interior angles are equal,

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exactly three sides of the trapezium are equal

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(C) Given that $AC$ is the bisector of $\angle DAB$,we have $\angle DAC = \angle CAB$.Since $DC \parallel AB$,the alternate interior angles are equal,so $\angle CAB = \angle ACD$.Therefore,$\angle DAC = \angle ACD$,which implies that $	riangle ADC$ is an isosceles triangle with $AD = DC$.Similarly,since $BD$ is the bisector of $\angle CBA$,we have $\angle CBD = \angle DBA$.Since $DC \parallel AB$,the alternate interior angles are equal,so $\angle DBA = \angle BDC$.Therefore,$\angle CBD = \angle BDC$,which implies that $	riangle BCD$ is an isosceles triangle with $BC = DC$.Since $AD = DC$ and $BC = DC$,we have $AD = BC = DC$.Thus,exactly three sides of the trapezium are equal.

Suppose $ABCD$ $(AB \parallel CD)$ is a trapezium such that the diagonals $AC$ and $BD$ bisect the angles $\angle DAB$ and $\angle CBA$,respectively. Then

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