Question

please answer me question no 25??

Answer 1

#RAM RAM ji ❤^_^

ĀNSWĒR ⏬⏬

$given = quadrilateral \: ABCD \\ \\ in \: which \: ∠B = 90° \: and \\ \\ { AD}^{2} = {AB}^{2} + {BC}^{2} + {CD}^{2} \\ \\ To \: prove = ∠ACD = 90° \: \\ \\ Proof \: In \: right \: angle \: triangle \: \\ ABC \: \\ \\ {AC}^{2} = {AB}^{2} + {BC}^{2} (by \: Pythagoras \: \\ theorem) (equation1) \\ given = \\ {AB}^{2} + {BC}^{2} + {CD}^{2} \: \ \\ = \ { AD}^{2} = {AC}^{2} \: + {CD}^{2} \: \\ (by \: using \: equation1.) \\ \\ now.in \: triangle \: ACD \: \\ = { AD}^{2} = {AC}^{2} + {CD}^{2} \: \: \\ ∴ ( by \ Converse of \\ Pythagoras \: theorem) \\ ∴ ∠ACD = 90°$

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