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2015-05-09T13:17:22+05:30

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GIVEN ;-

⇒ E is the mid point of parallelogram of  ABCD.

⇒ AB and CE bisects angle BCD.

CONSTRUCTION :-

⇒Draw EF parallel to AD in parallelogram ABCD

TO PROVE ;-

⇒i) AE  =  AD

ii) DE bisects angle ADC 

iii) Angle DEC is a right angle

PROOF :-

⇒ In the question it is given that,  E is the mid-point of AB in Parallelogram ABCD .

⇒ Now In parallelogram ABCD,

  ⇒      ∠BCE = ∠DCE    {Because  ⇒CE is  the bisector of ∠BCD}.

⇒          BE = BC            [ Because ,Opposite sides of equal angles are equal.]

⇒        ∠DCE = ∠BEC     { This both angles are alternative angles}

⇒           AE   =   AD        { In parallelogram ABCD, E is the midpoint of AB,                                            BC and AD and this are opp. sides of parallelogram }
 

If AD and AE are equal then we get as ,

       ∠ADE = ∠AED           [because Opposite angles of equal sides are equal.]



⇒  But, ∠AED = ∠EDC. [ This both. are .Alternate angles]

so ,

⇒         ∠ADE = ∠EDC      [  As , DE is the bisector of ∠D ]



Let us take the given three angle as x, we get as ,

               ⇒ ∠ADE = ∠AED = ∠CDE = x

Let us take the given three angles as y , we get as,

               ⇒ ∠BCE = ∠BEC = ∠DCE = y

We know that ,

                             ⇒∠DEF = x [Alternate angles in parallelogram ABCD]

                             ⇒∠CEF = y [Alternate angles in parallelogram ABCD]

And , In ∠AEB ,

∠AEB   =    x    +     x    +   y     +    y    =    180°


                     2  (   x +  y   )                    = 180°


                           ( x + y)                        =  180° \  2

   
                            (x + y)                        =   90°



Hence it is 90 degree so it is right angle .

so, ∠DEC is a right angle

Hence proved.


 

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