Question

[tex]\huge \mathfrak \red{Question}Question

Q.1: ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that

(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC.​

Answer 1

$\bold \color{green}{ \fcolorbox{blue}{purple}{required \: answer}}$

$\bold \color{green}{ \underline{ \underline \red{given \tt}}}$

AD = BC

$\angle \: DAB= \angle \: CBA$

$\bold \color{green} { \underline{ \underline \red{to \: prove \tt}}}$

i)∆ABD = ∆BAC

$\bold { \underline{ \underline{proof}}}$

$\bold{in \: \triangle ABD \: \: and \: \triangle BAC}$

$AB = AD(given) \: \: \: \: \: \: \: \: \: \: \: \\ \angle \: DAB = \angle \: CBA(given) \\ AB= AB(common) \: \: \: \: \: \\ \\ \bold{ \implies \triangle \: ABD \cong \triangle \: BAC } \\ by \: SAS \: congruency \: rule$

ii) BD = AD

$since, \: \triangle \: ABD \cong \triangle \: BAC \\ \implies \: BD \: = AC \: ( \: by \: C.P.C.T)$

iii)

$\huge \bold \angle \: ABD = \angle \: BAC$

$\bold{ \underline{ \underline{proof}}}$

$since, \: \triangle \: ABD \cong \triangle \: BAC \\ \implies \angle \: ABD \: = \angle \: BAC \: ( \: by \: C.P.C.T)$

Answer 2

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