1.

Let ABCD be a quadrilateral with /_CBD=2/_ADB, /_ABD=2/_CDB, AB=BC, then

Answer»

`AD=CD`
`/_ADB=/_CDB`
`/_CBD=/_ABD`
`/_ADC` is`(2pi)/3`

Solution :`2x=/_CBD`
`2y=/_ABD`
In `/_\CBD`
`(sin(pi-(2y+x)))/(sinx)=(BD)/(BA)=(BD)/(BC)=(sin(pi-(2x+y)))/(SINY)`
`impliessin(2+x)siny=sin(2x+y)sinx`
`=1/2[cos(y+x)-cos(3y+x)]=1/2[cos(x+y)-cos(3x+y)]`
`0ltx+y=1/2ABClt(pi)/2`
`0LT(3y+x)+(3x+y)lt2pi`
`=:. 3y+x=3x+yimpliesx=y`
`implies/_ABD=/_CBDimpliesAD=CD`


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