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In the adjoining figure, $$ABCD$$ is a quadrilateral. The bisectors of $$\angle B$$ and the exterior angle at $$D$$ meet at $$P$$. Given $$\angle C = 80^\circ$$, $$\angle ADC = \frac{1}{2}\angle A$$ and $$\angle A = \angle C + 40^\circ$$. Then $$\angle DPB$$ is
Here $$\angle A = 80^\circ+40^\circ=120^\circ$$ and $$\angle ADC = \frac{1}{2}\angle A = 60^\circ$$, so by the angle sum of the quadrilateral $$\angle B = 360^\circ-120^\circ-80^\circ-60^\circ=100^\circ$$. Chasing angles along the bisector of $$\angle B$$ and the bisector of the $$120^\circ$$ exterior angle at $$D$$ down to their meeting point $$P$$ gives $$\angle DPB = 70^\circ$$.
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