in the following figure ABCD is a parallelogram prove that AP bisects angle A B P bisects Angle B angle d a p + angle cbp equals to angle APB
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∠APB = ∠DAP + ∠CBP if ABCD is a parallelogram and AP & BP bisects ∠A & ∠B
Step-by-step explanation:
in a parallelogram
sum of adjacent angles = 180°
=> ∠A + ∠B = 180°
AP bisects ∠A
=> ∠BAP = ∠A - ∠DAP
BP bisects ∠B
=> ∠ABP = ∠B - ∠CBP
in Δ ABP
∠BAP + ∠ABP + ∠APB = 180°
∠A - ∠DAP + ∠B - ∠CBP + ∠APB = 180°
=> ( ∠A + ∠B) + ∠APB = 180° + ∠DAP + ∠CBP
=> 180° + ∠APB = 180° + ∠DAP + ∠CBP
=> ∠APB = ∠DAP + ∠CBP
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