Question

R is the midpoint of the side BC of parallelogram ABCD and AE is the bisector of Prove that 2AB= AD

Answer 1

parallelogram ABCD, E is the mid-point of AB. CE is the bisector of ∠BCD. ⇒ ∠BCE = ∠DCE ⇒ ∠DCE = ∠BEC [Alternate angles] ⇒ BE = BC [Opposite sides of equal angles are equal.] ⇒ AE = AD [E is the midpoint of AB, BC and AD are opposite sides of a parallelogram.] AE and AD are equal. Therefore, ∠ADE = ∠AED [Opposite angles of equal sides are equal.] But, ∠AED = ∠EDC. [Alternate angles] ⇒ ∠ADE = ∠EDC ⇒ DE is the bisector of ∠D 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

Answer 2

$\mathbb{\pink{\huge{Hello\:Mate!!}}}$

$<b><u>Answer</u></b>$

$<b>To prove => 2AB = AD</b>$

$<b><u>Step-by-step explanation</u></b>$

Let DE cut AB extended at F (See the figure).

$<b>We have,</b>$

∠BEF = ∠DEC

△FBC = △ECD

So, △CED ≅ △BEF (Case ASA)

So, EF = ED and CD = BF = AB

$<b>In △AFF,</b>$

AE is the median and angle bisector of ∠A.

So, △ADF is isosceles.

So, AD = AF = 2

AB => AB = $\frac{1}{2}$ AD

$<b>(or)</b>$

$<b>2AB = AD</b>$

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