Angles CAB and BDC, being both inscribed angles intercepting the same arc (BC) are equal.
Let the measure of arc BC be 2a degrees, then we know that the inscribed angles CAB and BDC have a measure of a degrees.
Similarly, let the measure of arc AD be 2b degrees, then the measures of the angles ACD and ABD are b degrees.
The sum of the measures of arcs BC and AD, 2a+2b, is:
is 360°-m arc(BD) -m arc(CA)=360°-170°-70°= 120°, which is the answer of the second question.
We also note that a+b is half of 120°, that is 60°.
Note that angle BPC is an exterior angle of triangle APC, so the measure of this angle is equal to the sum of the two opposing interior angles, which is a+b.
Thus, the measure of angle BPC is 60°, which means that the measure of angle BPD is 120°, since these two angles are supplementary.
Answers:
i) 120°
ii) 60°
