Question

in the figure given a long side ABCD is a rombus such that AP=AB=BQ . PD & QC are produced to meet at a point R . show that angle PRQ =90°

Answer 1

Given PA = AB = BQ,
We know AB = CD = BC = AD   (sides of rhombus are equals)
And ∠DOC = ∠AOD = ∠AOB = ∠BOC = 90°   ( Diagonal of rhombus are perpendicular to each other)
In ∆PAD, We know
  PA = AD 
So,
∠APD = ∠ADP = x    (Angle opposites same side are equal As PA = AD)
Then 
∠PAD = 180° - x
Similarly In ​ ∆BCQ
  BC = BQ 
So,
∠BCQ = ∠BQC = y   (Angle opposites same side are equal As PA = AD)
Then 
∠QBC = 180° - y
So, In  ∆ RDC
     ∠RDC = ∠APD = x          (As we know Two lines are parallel if and only if the two angles of any pair of corresponding angles of any transversal are congruent)
Therefore 
     ∠BQC = ∠RCD = y
And,
   ∠PAD = ∠ADC = 180° - x    (As we know Two lines are parallel if and only if the two angles of any pair of alternating angles of any transversal are congruent)
Therefore 
 ∠QBC​ = ∠BCD = 180° - y 
Now,
 ∠CDO = 1/2 ∠ADC = 90° - x  ( As we know diagonal of rhombus bisect its angles)
Therefore
∠DCO = 1/2 ∠BCD = 90° - y 
So In quadrilateral RDOC
 ∠DOC = 90°  
 ∠RDO = ∠CDO + ∠RDC 
            = 90° - x + x
           = 90°
∠RCO = ​∠DCO + ​∠RCD
           = 90° - y + y
          = 90°
SO the remainig ​∠DRC = 90°     (Because sum of internal angles of any quadrilateral is 180°)
Therefor 
∠PRQ = 90°              (Because ∠PRQ = ​∠DRC  )                          
Hence ​∠PRQ is a right angle                                                   (Hence proved)

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