Answer :
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.
⇒ 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.
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.
