__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.**