In ∆ABC and ∆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 :
1) quadrilateral ABED is a parallelogram.
2) quadrilateral BEFC is a parallelogram.
3) AD||CF and AD = CF.
4) quadrilateral ACFD is a parallelogram.
5) AC = DF.
6) ∆ABC = ∆DEF.
• no spam plz
• wrong answer will be reported
Answers & Comments
Verified answer
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]