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M manish painkra
We have,  PQR is produced to a point S and  bisectors of PQR and PRS meet at point T, By exterior angle sum property, PRS = P + PQR Now,  ................(i) Since QT and QR are the bisectors of  PQR and PRS respectively. Now, in QRT, ..............(ii) From eq (i) and eq (ii),  we get Hence proved

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M manish painkra
We have,  PQ PS, PQ || SR, SQR = 28° and QRT = 65° Now, In  QRS, the side SR produced to T and PQ || RS therefore, QRT =  =   So,  Also, QRT = RSQ + SQR (By exterior angle property of a triangle) Therefore, RSQ = QRT - SQR                     Now, in  PQS, P + PQS + PSQ =   

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M manish painkra
We have, lines PQ and RS intersect at point T, such that PRT = 40°, RPT = 95° and TSQ = 75° In PRT, by using angle sum property PRT + PTR + TPR =  So, PTR  =     Since lines, PQ and RS intersect at point T therefore, PTR = QTS (Vertically opposite angles)                 QTS =  Now, in QTS, By using angle sum property TSQ + STQ + SQT =  So, SQT = 

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M manish painkra
We have,  AB || DE,  BAC = 35° and  CDE = 53° AE is a transversal so,  BAC =  AED =  Now, In  CDE, CDE + DEC + ECD =  (By angle sum property) Therefore, ECD =                   

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M manish painkra
We have  X = , XYZ =  YO and ZO bisects the XYZ and XZY Now, In XYZ, by using angle sum property XYZ + YZX + ZXY =  So, YZX =        YZX =  and, OYZ =  also, OZY =  Now, in OYZ  Y + O + Z =   [Y =  and Z = ] So, YOZ = 

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M manish painkra
Given, PQR is a triangle, SPR =,  PQT =  Now, TQP + PQR =  (Linear pair) So, PQR =  Since the side of QP of the triangle, PQR is produced to S So, PQR + PRQ =  (Exterior angle property of triangle)

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M manish painkra
Draw  a ray BL PQ and CM  RS Since PQ || RS (Given) So, BL || CM and BC is a transversal  LBC =  MCB (Alternate interior angles).............(i) It is known that, angle of incidence  = angle of reflection So, ABL = LBC and MCB =  MCD ..................(ii) Adding eq (i) and eq (ii), we get ABC = DCB Both the interior angles are equal Hence AB || CD

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M manish painkra
Given, AB || CD, APQ =  and PRD =  PQ is a transversal. So,  APQ = PQR=  (alternate interior angles) Again, PR is a transversal. So,  y + =  (Alternate interior angles)

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M manish painkra
Draw a line EF parallel to the ST through R. Since PQ || ST and ST || EF   EF || PQ PQR = QRF =   (Alternate interior angles) QRF = QRS + SRF .............(i) Again, RST + SRF =  (Interior angles of two parallels ST and RF)   (RST = , given) Thus, QRS = 

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M manish painkra
Given AB || CD, EFCD and GED =  In the above figure,  GE is transversal. So, that AGE = GED =   [Alternate interior angles] Also, GEF = GED - FED                       =            Since AB is a straight line  Therefore, AGE  + FGE =   So, FGE = 

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M manish painkra
Given AB || CD and CD || EF and  therefore, AB || EF and  (alternate interior angles)..............(i) Again, CD || AB  .............(ii) Put the value of  in equation (ii), we get Then  By equation (i), we get the value of   

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M manish painkra
Given that, In the figure, CD and PQ intersect at  F Therefore,   (vertically opposite angles) PQ is a straight line. So,  Hence AB || CD (since  and are  alternate interior angles)

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M manish painkra
Given that, XYZ =  and XY produced to point P and Ray YQ bisects ZYP   Now, XYP is a straight line So, XYZ + ZYQ + QYP =  Thus reflex of QYP =  Since XYQ = XYZ + ZYQ  [      = 

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M manish painkra
Given that, POQ is a line, OR  PQ and  ROQ is a right angle. Now,  POS + ROS +  ROQ =   [since POQ is a straight line]  .............(i) and,  ROS +  ROQ =  QOS        ..............(ii) Add the eq (i ) and eq (ii),  we get hence proved

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M manish painkra
Given that,  ..............(i) It is known that, the sum of all the angles at a point =    ..............(ii) From eq (i) and eq (ii), we get Hence proved AOB is a line.  

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M manish painkra
Given that, ABC is a triangle such that  PQR =  PRQ  and ST is a straight line. Now,  PQR +  PQS =      {Linear pair}............(i) Similarly,  PRQ +  PRT = ..................(ii) equating the eq (i) and eq (ii), we get    {but  PQR =  PRQ } Therefore,  PQS = PRT Hence proved.

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M manish painkra
Given that, Line XY and MN intersect at O and POY =  also ..............(i)  Since XY is a straight line Therefore, ...........(ii) Thus, from eq (i) and eq (ii), we get So,   Since MOY = c [vertically opposite angles]           a + POY = c            

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M manish painkra
Given that, AB is a straight line. Lines AB and CD intersect at O.  and  BOD =  Since AB is a straight line  AOC + COE + EOB =   [since ] So, reflex COE =  It is given that AB and CD intersect at O Therefore, AOC  = BOD  [vertically opposite angle]  [ GIven  BOD = ] Also,  So,  BOE = 

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D Devendra Khairwa
(i) In this case. the sum of the interior angle is  thus l is not parallel to m.\ (ii) In this case also l is not parallel to m as the corresponding angle cannot be   (Linear pair will not form). (iii) In this l and m are parallel. This is because the corresponding angle is  and it forms linear pair with   . (iv) The lines are not parallel as the linear pair not form. (Since corresponding angle...

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S Sanket Gandhi
(i) Since side AB is parallel to DG.       Thus :                           (Corresponding angles of parallel arms are equal.)   (ii)     Further side BC is parallel to EF. We have :                             (Corresponding angles of parallel arms are equal.)
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