Math, asked by agarwalseema593, 2 months ago

11. In A ABC and A DEF, AB = DE, AB || DE, BC = EF
and BC || EF. Vertices A, B and C are joined to
vertices D, E and F respectively (see Fig. 8.22).
Show that
(1) quadrilateral ABED is a parallelogram
(ii) quadrilateral BEFC is a parallelogram
(ii) AD || CF and AD=CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC=DF
(vi) A ABC = A DEF
Fig. 8.22​

Answers

Answered by mathdude500
2

\sf 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

\large\underline{\sf{Solution-}}

Given that,

\sf \: AB \:  \parallel \:DE \: and \: AB = DE \\  \\

We know,

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

\sf\implies \bf \: ABED \: is \: a \: parallelogram. \\  \\

\sf\implies\sf \: AD\:  \parallel \:BE \:  \: and \:  \: AD = BE -  -  - (1) \\  \\

Further given that,

\sf \: BC\:  \parallel \:EF \: and \: BC = EF \\  \\

We know,

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

\sf\implies \bf \: BEFC \: is \: a \: parallelogram. \\  \\

\sf\implies \sf \: BE\:  \parallel \:CF \:  \: and \:  \: BE = CF -  -  - (2) \\  \\

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

\sf\implies \bf \: AD\:  \parallel \:CF \:  \: and \:  \: AD = CF \\  \\

We know,

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

\sf\implies \bf \: ACFD \: is \: a \: parallelogram. \\  \\

\sf\implies \sf \: AC\:  \parallel \:DF \:  \: and \:  \: AC = DF \\  \\

Now,

\sf \: In \:  \triangle \: ABC \: and \:  \triangle \: DEF \\  \\

\qquad\sf \: AB = DE \:  \:  \:  \:  \{given \} \\  \\

\qquad\sf \: BC = EF \:  \:  \:  \:  \{given \} \\  \\

\qquad\sf \: AC = DF \:  \:  \:  \:  \{proved \: above \} \\  \\

\sf\implies \bf \:   \triangle \: ABC \:  \cong \:  \triangle \: DEF \:  \:  \:  \{SSS \: Congruency \: rule \} \\  \\

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