Answer
$\angle$ABC=35$^{\circ}$
$\angle$ADC=35$^{\circ}$
$\angle$1=110$^{\circ}$
Work Step by Step
In concave quadrilateral ABCD $\angle$A=40$^{\circ}$
$\triangle$ABD is isosceles. That means, $\angle$ABD=$\angle$ADB.
BC bisects $\angle$ABD which means, $\angle$ABC=$\angle$CBD
Also, DC bisects $\angle$ADB which means, $\angle$ADC=$\angle$CDB
Lets assume $\angle$ABD=$\angle$ADB=x
so $\angle$ABD+$\angle$ADB=2x
$\angle$A+$\angle$ABD+$\angle$ADB=180$^{\circ}$
40$^{\circ}$+2x=180$^{\circ}$
2x=180$^{\circ}$-40$^{\circ}$
x=$\frac{140}{2}$
x=70$^{\circ}$
$\angle$ABD=$\angle$ADB=70$^{\circ}$
$\angle$ABC=$\angle$CBD=$\angle$ABD$\div$2
$\angle$ABC=$\angle$CBD=35$^{\circ}$
$\angle$ABC=$\angle$ADC=35$^{\circ}$
now $\angle$1+$\angle$CBD+$\angle$CDB=180$^{\circ}$
$\angle$1+35$^{\circ}$+35$^{\circ}$=180$^{\circ}$
$\angle$1=180$^{\circ}$-70$^{\circ}$
$\angle$1=110$^{\circ}$
