Part (i)
We're told that [tex]\angle A = 60^{\circ}[/tex] which means angle DAB is 60 degrees. This angle is cut into two equal halves of 30 degrees each (for angles DAP and PAB) due to segment AP being a bisector of angle DAB.
Since ABCD is a parallelogram, the adjacent angles A and B are supplementary. A = 60 leads to...
A+B = 180
B = 180 - A
B = 180-60
B = 120
Angle ABC is 120 degrees which splits in half to get 60. The angles PBA and PBC are 60 degrees each.
Focus on triangle PAB. We found A = 30 and B = 60 earlier. That must mean:
P+A+B = 180
P+30+60 = 180
P+90 = 180
P = 180-90
P = 90
Triangle PAB is a right triangle with angle P being the 90 degree angle. This is another way of saying angle APB is 90 degrees.
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Part (ii)
From the previous part, we know that angles DAP and PAB are 30 degrees each. The alternate interior angles DPA and PAB are equal because we have a parallelogram, so angle DPA is also 30 degrees.
For triangle DPA, the base angles A and P are congruent (30 degrees each). This leads immediately to the fact triangle DPA is isosceles. The congruent sides are opposite the congruent base angles.
Therefore, AD = DP
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Also from part (i), we found that angles PBA and PBC were 60 degrees each.
Since we have a parallelogram, the alternate interior angles PBA and BPC are congruent, meaning that angle BPC is also 60.
For triangle PBC, we have the interior angles of:
P+B+C = 180
60+60+x = 180
120+x = 180
x = 180-120
x = 60
Each interior angle of triangle PBC is 60 degrees. Triangle PBC is equilateral. By definition, equilateral triangles have all sides the same length.
Therefore, PB = PC = BC.
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Part (iii)
ABCD is a parallelogram with congruent opposite sides.
Let AD = BC = x
From part (ii), we found that AD and DP were the same length. So DP = x as well.
Also from part (ii), we found that BC = PC, so PC = x
We can then say:
DC = DP + PC
DC = x + x
DC = 2x
DC = 2*AD
