SSC CGL 10 Aug 2017 Shift 1 Paper

$\mathrm{AD}=\frac{\mathrm{BC}}{2}$ ∴ AD =BD = DC
In ΔADC,
∠ACD = 40°
∠DAC = ∠ACD (AD = DC)
∴ ∠DAC = 40°
⇒ ∠ADB = ∠DAC + ∠ACD
= 40° + 40°
∴ ∠ADB = 80°
In ΔABD,
∵ AD = BD
∵ ∠BAD = ∠ABD
⇒ ∠DAB $=\frac{180°-80°}{2}=\frac{100°}{2}$
∴ ∠DAB = 50°