Appropriate Question :-

In Δ ABC and Δ DEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively. Show that

(i) quadrilateral ABED is a parallelogram.

(ii) quadrilateral BEFC is a parallelogram.

(iii) AD || CF and AD = CF.

(iv) quadrilateral ACFD is a parallelogram.

(v) AC = DF

(vi) Δ ABC ≅ Δ DEF

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that,

[tex]\sf \: AB \: \parallel \:DE \: and \: AB = DE \\ \\ [/tex]

We know,

In a quadrilateral, if one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram.

[tex]\sf\implies \bf \: ABED \: is \: a \: parallelogram. \\ \\ [/tex]

[tex]\sf\implies\sf \: AD\: \parallel \:BE \: \: and \: \: AD = BE - - - (1) \\ \\ [/tex]

Further given that,

[tex]\sf \: BC\: \parallel \:EF \: and \: BC = EF \\ \\ [/tex]

We know,

In a quadrilateral, if one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram.

[tex]\sf\implies \bf \: BEFC \: is \: a \: parallelogram. \\ \\ [/tex]

[tex]\sf\implies \sf \: BE\: \parallel \:CF \: \: and \: \: BE = CF - - - (2) \\ \\ [/tex]

From equation (1) and (2), we concluded that

[tex]\sf\implies \bf \: AD\: \parallel \:CF \: \: and \: \: AD = CF \\ \\ [/tex]

We know,

In a quadrilateral, if one pair of opposite sides are equal and parallel, then quadrilateral is a parallelogram.

[tex]\sf\implies \bf \: ACFD \: is \: a \: parallelogram. \\ \\ [/tex]

[tex]\sf\implies \sf \: AC\: \parallel \:DF \: \: and \: \: AC = DF \\ \\ [/tex]

Now,

[tex]\sf \: In \: \triangle \: ABC \: and \: \triangle \: DEF \\ \\ [/tex]

[tex]\qquad\sf \: AB = DE \: \: \: \: \{given \} \\ \\ [/tex]

[tex]\qquad\sf \: BC = EF \: \: \: \: \{given \} \\ \\ [/tex]

[tex]\qquad\sf \: AC = DF \: \: \: \: \{proved \: above \} \\ \\ [/tex]

[tex]\sf\implies \bf \: \triangle \: ABC \: \cong \: \triangle \: DEF \: \: \: \{SSS \: Congruency \: rule \} \\ \\ [/tex]

Verified answer

[tex] \rule{200pt}{4pt}[/tex]

(i) Quadrilateral ABED is a parallelogram

[tex] \; \star \; {\underline{\boxed{\pmb{\orange{\frak{ \; solution \; :- }}}}}}[/tex]

We have ,

[tex] \large\sf \implies[/tex] AB = DE (given) and

[tex] \large\sf \implies[/tex] AB || DE ( given)

ABED is a Quadrilateral in which a pair of opposite sides that are AB and DE are parallel and of equal length

Therefore, ABED is a parallelogram

[tex]\rule{200pt}{4pt}[/tex]

(ii) Quadrilateral BEFC is a parallelogram

[tex] \; \star \; {\underline{\boxed{\pmb{\orange{\frak{ \; solution\; :- }}}}}}[/tex]

We have ,

[tex] \large\sf\implies[/tex] BC = EF (given) and

[tex] \large\sf\implies[/tex] BC || EF ( given)

BEFC is a Quadrilateral in which a pair of opposite sides that are BC and EF are parallel and of equal length

Therefore, BEFC is a parallelogram

[tex]\rule{200pt}{4pt}[/tex]

(iii) AD || CF and AD = CF

[tex] \; \star \; {\underline{\boxed{\pmb{\orange{\frak{ \; solution \; :- }}}}}}[/tex]

ABED is a parallelogram

Therefore,

[tex] \large\sf \implies[/tex] AD || BE and AD = BE... (1)

Also, BEFC is a parallelogram

Therefore,

[tex] \large\sf \implies[/tex] BE || FC and BE = FC... (2)

From (1) and (2) , we have

AD || CF and AD = CF

[tex]\rule{200pt}{4pt}[/tex]

(iv) Quadrilateral AFCD is a parallelogram

[tex] \; \star \; {\underline{\boxed{\pmb{\orange{\frak{ \; solution \; :- }}}}}} [/tex]

In Quadrilateral AFCD , one pair of opposite sides (AD and CF as proved above) are parallel and of equal length

Therefore, Quadrilateral AFCD is a parallelogram

[tex]\rule{200pt}{4pt}[/tex]

(v) AC = DF

[tex] \; \star \; {\underline{\boxed{\pmb{\orange{\frak{ \; solution \; :- }}}}}} [/tex]

Since , AFCD is a parallelogram

So, AC = DF (because these are opposite sides of parallelogram)

[tex]\rule{200pt}{4pt}[/tex]

(vi) ∆ABC = ∆DEF

[tex] \; \star \; {\underline{\boxed{\pmb{\orange{\frak{ \; solution \; :- }}}}}}[/tex]

In ∆ABC and ∆DEF

AB = DE

BC = EF

AC = DE

Therefore, ∆ABC = ∆DEF

[tex]\rule{200pt}{4pt}[/tex]