Question

can anyone can do this for me

Answer 1

(i) AD=BC (Opp. sides of a II gram are equal)
Angle QBC=Angle PDA (Alternate angles)
DP=BQ (Given) .
  .
.   . Triangles APD and CQB are congruent to each other. (By SAS)

(ii) AP=CQ [Congruent Parts of Congruent Triangles(APD & CQB) are equal]

(iii) AB=CD (Opp. sides of a II gram are equal)
Angle ABQ=Angle PDC (Alternate angles)
DP=BQ (Given)
 .
.  . Triangles AQB and CPD are congruent to each other. (By SAS)

(iv) AQ=CP[Congruent Parts of Congruent Triangles(AQB & CPD) are Equal]

(v) From above proofs we can say :
AQ=CP
AP=CQ
Now,
QP=PQ(Common side)
 .
.  . Triangles AQP &CPQ are congruent to each other.(By SSS)
So, Angle APQ=Angle PQC.[Congruent Parts of Congruent Triangles(AQP & CPQ) are Equal]
Hence, AP II CQ (Angles APQ &PQC are alternate)
Also, Angles AQP=Angle PCQ [Congruent Parts of Congruent Triangles(AQP & CPQ) are Equal]
So, AQ IICP (Angles AQP & PCQ are alternate)
We know from earlier proofs that :
AP=CQ & AQ=CP.
Hence, in quadrilateral APCQ, opp.sides are equal and parallel.
This proves that APCQ is a parallelogram.