Q : 6     ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if
             necessary) at E. Prove that \small AE=AD.

Answers (1)

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove :   AE = AD 

Proof : 

              

          \angleADC = \angle3  , \angleABC = \angle4, \angleADE = \angle1  and \angleAED = \angle2

           \angle 3+\angle 1=180 \degree.................1(linear pair)

         \angle 2+\angle 4=180 \degree....................2(sum of opposite angles of cyclic quadrilateral)

         \angle3 = \angle4      (oppsoite angles of parallelogram )

 From 1 and 2,

           \angle3+\angle1 = \angle2 + \angle

From 3,    \angle1 = \angle

From 4,    \triangleAQB,     \angle1 = \angle2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

 

 

 

 

 

  

 

 

 

 

 

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