(i) We know that opposite angles are equal in a parallelogram.

So we get

∠ C = ∠ A = 60^o

We know that the sum of all the angles in a parallelogram is 360^o

It can be written as

∠ A + ∠ B + ∠ C + ∠ D = 360^o

So we get

∠ B + ∠ D = 360^o – (∠ A + ∠ C)

By substituting values in the above equation we get

∠ B + ∠ D = 360^o – (60^o + 60^o)

On further calculation we get

∠ B + ∠ D = 360^o – 120^o

By subtraction

∠ B + ∠ D = 240^o

We know that ∠ B = ∠ B

So the above equation becomes

∠ B + ∠ B = 240^o

2 ∠ B = 240^o

By division

∠ B = ∠ D = 120^o

We know that AB || DP and AP is a transversal

From the figure we know that ∠ APD and ∠ PAD are alternate angles

∠ APD = ∠ PAD = 60^o/2

∠ APD = ∠ PAD = 30^o ……. (1)

We know that AB || PC and BP is a transversal

So we get

∠ ABP = ∠ CPB = ∠ B/2

i.e. ∠ ABP = ∠ CPB = 120^o/2

We get ∠ ABP = ∠ CPB = 60^o …….. (2)

We know that DPC is a straight line

It can be written as

∠ APD + ∠ APB + ∠ CPB = 180^o

By substituting the values we get

30^o + ∠ APB + 60^o = 180^o

On further calculation

∠ APB = 180^o – 30^o – 60^o

By subtraction

∠ APB = 180^o – 90^o

∠ APB = 90^o

Therefore, it is proved that ∠ APB = 90^o

(ii) From equation (1) we know that

∠ APD = 30^o

We know that

∠ DAP = 60^o/2

By division

∠ DAP = 30^o

So we get

∠ APD = ∠ DAP ……… (3)

From the figure we know that the sides of an isosceles triangle are equal

So we get

DP = AD

We know that

∠ CPB = 60^o and ∠ C = 60^o

By sum property of triangle

We get

∠ C + ∠ CPB + ∠ PBC = 180^o

By substituting the values in the above equation

60^o + 60^o + ∠ PBC = 180^o

On further calculation we get

∠ PBC = 180^o – 60^o – 60^o

By subtraction

∠ PBC = 180^o – 120^o

∠ PBC = 60^o

We know that all the sides of an equilateral triangle are equal

So we get

PB = PC = BC ………. (4)

Therefore, it is proved that AD = AP and PB = PC = BC.

(iii) We know that ∠ DPA = ∠ PAD based on equation (3)

We also know that all the sides are equal in an isosceles triangle

DP = AD

From the figure we know that the opposite sides are equal

So we get

DP = BC

Considering equation (4)

DP = PC

From the figure we know that DP = PC and P is the midpoint of the line DC

So we get

DP = ½ DC

By cross multiplication we get

DC = 2AD

Therefore, it is proved that DC = 2AD.