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Let ABCD be a quadrilateral with /CBD=2/...

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

A

`AD=CD`

B

`/_ADB=/_CDB`

C

`/_CBD=/_ABD`

D

`/_ADC` is`(2pi)/3`

Text Solution

Verified by Experts

The correct Answer is:
A, B, C
`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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