Question

ABCD is a rhombus whose side AB is produced to points P and Q such that AP=AB=BQ. PD and QC are produced to meet at a point R. Show that <PRQ = 90 degree

Answer 1

for diagram see the attachments below
given,
            PA=AB=BQ
we know that AB=CD=BC=AD
so angleDOC=angleAOD=AOB=BOC(diagonals bisect each other perpendicularly)
in ΔPAD, PA=PD
∧APD=∧ADP=x
so,
∠PAD=180-x
similarly in ΔBCQ ,
∠BCQ=∠BQC=y
so,
∠QBC=180-y
∠RDC=∠APD=x(corresponding angles r equal)
∠BQC=∠RCD=y(corresponding angles r equal)
∠PAD=∠ADC=180-x(corresponding angles r equal)
∠QBC=∠BCD=180-y
now,
∠CDO=1/2∠ADC=90-x
DCO=1/2∠BCD=90-y

so in quadrilateral RDOC,
∠DOC=90degree
∠RDO=∠CDO+∠RDC
         =90-x+x
          =90 degree
  
∠RCO=∠DCO+∠RCD
          =90-y+y
           = 90
          so yhe remaining is ∠DRC=90 degree(angle sum property of a quadrilateral)
∴∠PRQ=90 degree
∵∠PRQ=∠DRC
  
                Hence PRQ is  a right angled triangle
                    if helped plzzz mark it as the brainliest








                      

Answer 2

thats right answer thankyou