Question

Q7. ABCD is a parallelogram and X and Yare the mid-points of AB and CD resp. Show that AD∥XY. Q8.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°. Best answer will be marked as brainliest. Wrong answer will be reported.

Answer 1

Answer:

Step-by-step explanation:

           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 the remaining is ∠DRC=90 degree(angle sum property of a quadrilateral)

• ∴∠PRQ=90 degree

• ∵∠PRQ=∠DRC

 

              Hence PRQ is  a right angled triangle

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