# In parallelogram ABCD,E is the mid-point of side AB and CE bisects angle BCD.Prove that: i)AE=AD ii)DE bisects angle ADC iii)Angle DEC is a right angle. PLZ FAST WHOEVER WRITES IN 10 TO 15 MINS WOULD BE THE BEST ANSWERER.

1
by deodanChaudhary

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 ;-

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.