Question

Please answer this question as fast as possible and in full steps.

Answer 1

Take triangle CDQ and triangle PBC

We know,

AD = BC [ ABCD is a paralleogram​ ]

and QD = AD [ ∆ ADQ is a equilateral triangle]

So, QD = AD = BC

Again,

PB = AB= BC ( Triangle APB is a equilateral triangle and the opposite sides of paralleogram are equal)

Angle PBC = Angle QDC

So ∆ PBC =~ ∆ QDC ( according to SAS)

Now we have,

PC = QC

Now,

In ∆ PAQ and ∆ CQD

PA = CD [ as CD = AB and AB = PA]

AQ = DQ [ ∆ ADQ is a equilateral triangle]

Angle PAQ = Angle CDQ [ as PAQ = 60° + angle ABC and CDQ = 60° + angle ADC

ABC = ADC]

∆ PAQ =~ CQD

Now,

PQ = CQ

As, CQ = PC

PQ = CQ = PC

The triangle CPQ is an equilateral triangle.

Hope it helps.

Answer 2

let /_ABC = x°
CDA = ABC = x°
[opp angles of a llgm are equal]

/_BAD = 180 - x°
[corresponding angles are supplementary]

we know
angles of a equilateral triangle is 60.

/_QDC = x + 60
/_PBC = x + 60
/_QAP = 360 - (60+60+180-x)
= 360 - 300 + x
= x + 60

now
in triangles APQ , QDC, PBC
AQ= DC = PB [ given]
AP = QD= BC [given]
/_QAP = /_QDC = /_PBC [all are (x + 60)°]

therefore
APQ, QDC , PBC are congruent [bySAS]

PQ = QC = PC [cpct]
now PCQ is a triangle whose all the sides are equal .

therefore PCQ is an equilateral triangle