Question

HI BABES , SOLVE THIS FOR ME.....WILL MARK UH BRAINLIEST IF YOUR ANSWER IS RIGHT AND PROPER OTHERWISE I'LL REPORT NOT ONLY YOUR ANSWER BUT YOUR PROFILE.​

Answer 1

(i)

In parallelogram ABCD, E is the mid-point of AB.

CE is the bisector of ∠BCD.

⇒ ∠BCE = ∠DCE

⇒ ∠DCE = ∠BEC

⇒ BE = BC

⇒ AE = AD

(ii)

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

But, ∠AED = ∠EDC. [Alternate angles]

⇒ ∠ADE = ∠EDC

⇒ DE is the bisector of ∠D

(iii)

Mark the midpoint of CD as F.

Draw EF parallel to AD.

Let ∠ADE = ∠AED = ∠CDE = x

∠BCE = ∠BEC = ∠DCE = y

∠DEF = x [Alternate angles]

∠CEF = y [Alternate angles]

∠AEB = x + x + y + y = 180°

2(x + y) = 180°

(x + y) = 90°

Therefore, ∠DEC is a right angle

Hope this helps!

Answer 2

Since ABCD is a parallelogram, thenAB∥DC and AD∥BC.Now, CE bisects ∠DCB, so∠DCE = ∠BCE    ......(1)Since, DC∥AB and CE is a transversal, then∠DCE = ∠CEB   (Alternate interior angles)  .....(2)Now, from (1) and (2), we get∠BCE = ∠CEBIn ΔBCE,∠BCE = ∠CEB  (Proved above)⇒EB = BC  (Sides opposite to equal angles are equal)   ......(3)Since E is the mid point of AB, then AE = EB         .......(4)We know that opposite sides of ∥gm are equal, thenAD = BC       ......(5)Now, from (3), we get     EB = BC⇒AE = AD   [using (4) and (5)]In ΔDAE,     AE = AD  ⇒∠EDA = ∠AED   [Angles opposite to equal sides are equal]    ......(6)Since, DC∥AB and DE is a transversal, then∠EDC = ∠AED           (Alternate interior angles)  .....(7)From (6) and (7), we get∠EDA = ∠EDC⇒DE bisects ∠ADC.In ΔEDC,   ∠EDC + ∠DCE + ∠DEC = 180°   [Angle sum property]⇒12∠D + 12∠C + ∠DEC = 180°⇒∠DEC = 180° − 12(∠D + ∠C) ⇒∠DEC = 180° − 12×180°            [Adjacent angles of ∥gm are supplementary]⇒∠DEC = 90°