Question

in the fig AB = CD, <ABD =<CDB prove that AD=CB

Answer 1

Hey Friends!!

Here is your answer↓⬇

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$\huge \bf \sf{It \: is \: given \: that:-)}$

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$\bf{AB = CD.}$

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$\bf{ \angle \: ABD = \angle \: CDB .}$

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$\huge \bf \sf{To \: prove:-)}$

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$\huge \boxed{AD = C B .}$

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$\bf \underline{Now, \: In \: \triangle \: ABD \: and \: \triangle \: CDB.}$

$\bf{AB = CD ........(given).}$

$\bf{ \angle \: ABD = \angle \: CDB.......(given).}$

$\bf{BD = DB.........(common \: side).}$

$\boxed{ = > by \: SAS \: congruency \: rule}$

$\huge \bf{ = > \triangle ABD \cong \triangle CDB.}$

$\boxed{AD = CB.......(CPCT).}$

✅✅ Hence, it is proved ✔✔.

$\huge \boxed{THANKS}$

$\huge \bf \underline{Hope \: it \: is \: helpful \: for \: you}$

Answer 2

$\huge \bf{hey}$

$\huge \: here \: is \: the \: answer$
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$\huge{given \: }$
$\huge{ab = cd}$

$\huge{angle \: abd = angle \: cbd}$

$\huge{to \: prove \: ad = cb}$

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$\bf{in \: triangle \: abd \: and \: cbd}$

$\bf{ad = cb (given)}$

$\bf{angle \: abd = angle \: cbd(given) }$

$\bf{bd = bd(common)}$
hence

∆abd congruent ∆ cbd

hence

$\huge{ab = cd(cpct)}$
$\huge \: hope \: it \: helps$

$\huge \boxed{thanks}$