IN ∆ADB AD = AB Thus ∆ABC is an Isoceles Triangle So , Angle ADB = Angle ABD ( Isoceles Triangle property) angle A + angle ABD + angle ADB = 180° 130° + 2( angle ABD ) = 180° angle ABD = (180 - 130)/2 angle ABD = 25° angle ABD = angle ADB = 25° angle ADB = angle DBC (alternate interior angles) x = 25°
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IN ∆ADB
AD = AB
Thus ∆ABC is an Isoceles Triangle
So ,
Angle ADB = Angle ABD ( Isoceles Triangle property)
angle A + angle ABD + angle ADB = 180°
130° + 2( angle ABD ) = 180°
angle ABD = (180 - 130)/2
angle ABD = 25°
angle ABD = angle ADB = 25°
angle ADB = angle DBC (alternate interior angles)
x = 25°
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