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Let ABCD be a quadrilateral with /_CBD=2/_ADB, /_ABD=2/_CDB, AB=BC, then |
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Answer» `AD=CD` `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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