Question

$\huge \red{ \bold{ \underline{ \overline{ \mid{ \tt{question}} \mid}}}}$

in figure 5.13 ◻️ han hota hai ABCD is a parallelogram point is on the re AB such that b is equals to b then prove that line segment bc at point F.
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Answer 1

◻️ ABCD is a parallelogram

so,

AD || CB

$\displaystyle{ \implies{ \: ad \: || \: bf}}$

$\large{ \underline{Given}}$

AB = BE so, B is the midpoint of AE.

by the Converse of the midpoint theorem

F is the midpoint of DE.

so, DF = FE

in ∆ EBF and ∆ DCF,

DF = FE

${ \red{ \fbox{(proved \: above)}}}$

DC = BF

$\pink{(since \: dc \: = a \: b \: and \: a \: b \: = b \: e)}$

angel FDC = angel FEB

$\green{(alternate \: interior \: angel \: of \: parallel \: line \: a \: e \: and \: c \: d)}$

thus,

∆ ABF congruent ∆ DCF

$\orange{(s.a.s \: congruency)}$

$\huge{ \large{ \bold{ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: }}}}$

therefore, EB = FC (C.P.C.T)

Hence,

$\large{ \blue{ \fbox{ \tt{ed \: bisects \: seg \: bc \: at \: f}}}}$

hope its help u

$\huge{ \fbox{ army}}$

Answer 2

Answer:⋆

Answer Given:ABCD is a parallelogram .

Point E is on the ray AB such thatBE = AB .

To prove : BF = FCProof:i ) AB = DC [ opposite sides ]AB = BE ( given )DC = BE --- ( 1 )ii ) In ∆DCF and ∆EBF<FDC = <FEB [ alternate angles ]DC = BE ( S ) [ from ( 1 ) ]<DFC = <EFB [ veritically opposite angles ]Therefore ,∆DCF is congruent to ∆EBF[ ASA congruence Rule ]CF = FB

[ CPCTStep-by-step explanation:

hope so my answer is right

hii good morning sis