Question

ABCD is a parallelogram AB produced to E so that BE = AB prove that ad bisects BC


please answer the question..​

Answer 1

Step-by-step explanation:

given -ABCD IS A PARALLELOGRAM

BE=AB

TO SHOW-

ED BISECT BC

PROOF-

AB=BC(GIVEN)

AB=CD( OPPOSITE SIDE OF PARALLELOGRAM)

-BE=CD

LET DE INTERSECT BC AT AF.

NOW,

IN ∆CDO AND .∆BEO

angle DCO = angle EBO (AE ||CD)

angle DOC =angle EOB (VERTICALLY OPPOSITE ANGLE)

BE = CD PROVED.

∆CDO congruent ∆BEO ( AAS RULE)

THUS , BF = FC (CPCT)

AD BISECT BC.

Answer 2

$\huge \underline\textbf{Solution:-}$

$\green{\bold{Since\: ABCD\:is\:a\:parallelogram.}}$

$\green{\bold{Therefore,\:AB||DC}}$

(see above attachment)

Now, AB || DC and transversal BC intersects them .

∴ ⠀⠀⠀⠀⠀⠀∠1 = ∠2⠀⠀⠀⠀⠀...... (1)

$\green{\bold{Now,\:ABCD\:is\:a\:parallelogram}}$

=> ⠀⠀⠀⠀⠀AB = DC

=>⠀⠀ ⠀⠀⠀BE = DC⠀⠀⠀[∵ AB = BE (Given)] ...... (2)

$\green{\bold{Thus\:in \:Triangle\:BOE \: and\:COD, \:we\: have}}$

⠀⠀⠀⠀⠀⠀⠀∠1 = ∠2⠀⠀⠀⠀⠀⠀⠀[From (1)]

⠀⠀⠀⠀⠀⠀⠀∠3 = ∠4⠀⠀[Vertically oppo. angles]

and, ⠀⠀⠀⠀BE = DC⠀⠀⠀⠀⠀⠀[From (2)]

$\green{\bold{So,\:by\:AAS\:criterion\: of \:congruence\: we \:obtain}}$

⠀⠀⠀⠀△BOE ≈△COD

=> ⠀ ⠀⠀⠀BO = CO

=> ⠀⠀O is the mid point of BC ⠀⠀⠀⠀⠀⠀[C.P.C.T]

=> $\sf\small\underline\orange{ED\:bisects \:BC}$