По теореме о биссектрисе
AB/BC =AD/DC =14/6 =7/3
AB=7x, BC=3x
AB-BC =4x =12 => x=3
P(ABC) =AD+DC+AB+BC =20+10x =20+30 =50
По теореме о биссектрисе
AB/BC =AD/DC =14/6 =7/3
AB=7x, BC=3x
AB-BC =4x =12 => x=3
P(ABC) =AD+DC+AB+BC =20+10x =20+30 =50