Question

In the adjoining figure, ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB ​

Answer 1

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Correct Question:-

In the adjoining figure, ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced to meet at F, prove that AF = 2AB.

$\large{\underline{\underline{\sf{\purple{Required\:Answer:-}}}}}$

• Consider △ DEC and △ FEB

From the figure we know that ∠ DEC and ∠ FEB are vertically opposite angles.⠀

⠀⠀$\implies{\rm }$∠ DEC = ∠ FEB

⠀⠀$\implies{\rm }$∠ DCE and ∠ FBE

⠀⠀$\implies{\rm }$∠ DCE = ∠ FBE

It is Given that :-

⠀⠀$\implies{\rm }$CE = EB

By AAS congruence criterion

⠀⠀$\implies{\rm }$△DEC ≅ △ FEB

⠀⠀$\implies{\rm }$DC = FB (c. p. c. t)

⠀⠀$\implies{\rm }$From the figure

⠀⠀$\implies{\rm }$AF = AB + BF

ωє кησω тнαт :-

⠀⠀$\implies{\rm }$BF = DC and AB = DC

ѕσ ωє Gєт,

⠀⠀$\implies{\rm }$AF = AB + DC

⠀⠀$\implies{\rm }$AF = AB + AB

By addition:-

⠀⠀ $\implies{\rm }$AF = 2AB

"Therefore, it is proved that AF = 2AB."

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Answer 2

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${ \huge{ \underline{ \underline{ \underline{ \underline{ \frak{ \red{Correct \: Question : }}}}}}}}$

--》ABCD is a  parallelogram and E is the mid-point of side

BC. IF DE and AB when produced meet at F, prove that 

 AF = 2AB.

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${ \huge{ \underline{ \underline{ \underline{ \underline{ \frak{ \red{Required \: Answer : }}}}}}}}$

♧Given♧

--》ABCD is a parallelogram

--》E is the mid point of side BC

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♧To Prove♧

--》AF = 2AB

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♧Solution♧

--》△DCE and△BFE

--》any DEC = any BEF (vertically opp any)

--》EC=BE (E is the mid point)

--》∠DCB=∠EBF (alternate angle DC parallel to AF)

--》So △DCE congruent to △BFE

--》Therefore DC=BF ...(1)

--》now CD = AB (ABCD is a parallelogram)

So,

--》AF=AB+BF

--》=AB+DC from (1)

--》=AB+AB

--》=2AB

$\sf \boxed {AF = 2AB}$

♧Hence Proved..♧

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