Question

in the adjoining figure ABCD is a parallelogram,E is the mid point of AB and CE bisect angle BCD prove that: (i) AE = AD (ii) DE bisects angle ADE (III) angle DEC = 90°​

Answer 1

Step-by-step explanation:

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.