Question

ABCD is a rhombus,then prove that AB square+BC square+CD square+DA square=AC square+BD square​

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt ABCD\: is \:a\: rhombus,\:then\: prove\: that \\\tt {AB}^{2}+{BC}^{2}+{CD}^{2}+{DA}^{2}={AC}^{2}+{BD}^{2}$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\bf ABCD \:is\:a\:rhombus$
• $\bf BD \:and \:AC\:diagonals \:meet \:at \:a\:point\:O$

$\sf \bf {\boxed {\mathbb {TO\:PROVE}}}$

• $\bf {AB}^{2}+{BC}^{2}+{CD}^{2}+{DA}^{2}={AC}^{2}+{BD}^{2}$

$\sf \bf {\boxed {\mathbb {PROOF}}}$

${\pink{ \underline {\mathfrak { Diagonals \: of \: the \: rhombus \: bisect \: each \:other \: at \: right \: angles}}}}$

• ${\longrightarrow{\boxed {\sf {\angle AOB=\angle BOC=\angle COD=\angle DOA ={90}^{\circ}}}}}$

• $\longrightarrow {\boxed {\sf {BO=\dfrac{BD}{2}}}}-{\sf \big(1\big)}$

• $\longrightarrow {\boxed {\sf {AO=\dfrac{AC}{2}}}}-{\sf \big(2\big)}$

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${\pmb {In\:\triangle ABO}}$

$\bf \implies \angle AOB={90}^{\circ}$

${\underbrace {\overbrace {\orange {\bf {\pmb {By\: Pythagoras \:theorem}}}}}}$

$\bf \implies {AB}^{2}={BO}^{2}+{AO}^{2}$

$\bf \implies {AB}^{2}={\bigg(\dfrac{BD}{2}\bigg)}^{2}+{\bigg(\dfrac{AC}{2}\bigg)}^{2} \longrightarrow \Big(According \:to\:eq \:1\:and \:2\Big)$

$\bf \implies {AB}^{2}=\bigg(\dfrac{{BD}^{2}}{4}\bigg)+ \bigg(\dfrac{{AC}^{2}}{4}\bigg)$

$\bf \implies {AB}^{2}=\dfrac{{BD}^{2}+{AC}^{2}}{4}$

$\bf \implies 4{AB}^{2}={BD}^{2}+{AC}^{2}$

$\bf \implies {AB}^{2}+{AB}^{2}+{AB}^{2}+{AB}^{2}={BD}^{2}+{AC}^{2}$

$\bf \implies {AB}^{2}+{BC}^{2}+{CD}^{2}+{DA}^{2}={BD}^{2}+{AC}^{2}\longrightarrow \Big(\because AB=BC=CD=DA\Big)$

${\implies {\blue {\boxed {\boxed {\purple {\mathfrak {{AB}^{2}+{BC}^{2}+{CD}^{2}+{DA}^{2}={BD}^{2}+{AC}^{2}}}}}}}}$

${\red{\underline {\overline {\pmb {\sf {\pmb {{\therefore} Hence\:Proved}}}}}}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \bf \huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\sf Area\:of \:rhombus =\dfrac{d_1 \times d_2}{2}$

$\sf Perimeter\:of \:rhombus =4a$