Mathematics, 20.12.2021 07:20, genaro19

Pls help with steps also

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answered: dre8343

Part (i)

We're told that $\angle A = 60^{\circ}$ 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 = 60B = 60C = x = unknown for now

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.