1). В ΔABD: ∠ABD = ∠BAD; 2•∠ABD = 180 - ∠BDA (1)
B ΔBDC: ∠DBC = ∠BCD; 2•∠DBC = 180 - ∠BDC (2)
Из (1) и (2): 2•(∠ABD + ∠DBC) = 360 - (∠BDA + ∠BDC)
2•∠ABC = 360 - 180
∠ABC = 90°
2). ∠A/2 + ∠C/2 = 180 - 130 = 50° => ∠A + ∠C = 100° => ∠B = 80°
3). Так как ∠BAC + ∠ABC + ∠BCA = ∠DAC + ∠ADC + ∠DCA = 180°,
и ∠BAC = ∠BAD + ∠DAC; ∠BCA = ∠BCD + ∠DCA, то:
∠BAD + ∠DAC + ∠ABC + ∠BCD + ∠DCA = ∠DAC + ∠ADC + ∠DCA
∠BAD + ∠ABC + ∠BCD = ∠ADC => ∠ABC < ∠ADC