Math, asked by sumababu434, 5 months ago

B
Fig. 8.21
B
11. In A ABC and A DEF, AB = DE AB II DE, BC=EF
and BCH EF Vertices A, B and C are joined to
vertices D, E and F respectively (see Fig. 8.22).
Show that
(0 quadrilateral ABED is a parallelogram
quadrilateral BEFC is a parallelogram
(HD ADIICF and AD=CF
(iv) quadrilateral ACFD is a parallelogram
(v) AC=DF
(vi) AABC ADEE.
F
E

Answers

Answered by mathdude500

$\sf Appropriate Question: \\$

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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