Question

pls help fast sorry for bad handwriting ​

Answer 1

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${ \rm{ \large \underline{given}}}$

${ \rm{ABCD \: is \: a \: figure \: where \: AB \parallel CD \: and \: AC \parallel \: BD}}$

${ \rm{BC \: is \: the \: diagonal \: in \: the \: figure}}$

${ \rm{ \large \underline {to \: prove}}}$

${ \rm{(i) \triangle \: ABC \approx\triangle BCD}}$

${ \rm{(ii)AC = BD}}$

${ \rm{ \large \underline{solution}}}$

${ \rm{(i)let \: ABC \: and \: BCD \: are \: two \: triangles \: where}}$

${ \rm{CD =AB}}$

${ \rm{ \angle{BCD} = \angle{ABC}}}$

${ \rm{ CB = CB}}$

${ \rm{hence(byS.A.S) \triangle{ABC} \cong \triangle{BCD}(proved)}}$

${ \rm{(ii)\triangle{ABC} \cong \triangle{BCD} \: so}}$

${ \rm{AC = BD}}$

${ \rm{hence \: AC = BD}}$

Answer 2

$\underline{\sf{\red{Given}}}$

ABCD is a figure where AB || CD and AC || BD

BC is the diagonal in the figure

$\tt\large{\red{To \:prove}}$

1 ΔABC = ΔBCD

2 . AC = BD

$\textbf{\large{\red{Solution}} }$

1 . let ABC and BCD are two triangle , where ,

CD = AB

angle C= angleB

CB = CB

hence by S.A.S Δ ABC = BCD ( proved )

so

AB = BD

$\tt\large{\red{hence \: AC = BD}}$

$\tt\large{\red{Proved}}$