Question

Given : □ ABCD is a Rhombus, Prove : 4AB^2 = AC^2 + BD^2
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Answer 1

$</p><p>Rhombus \: properties :</p><p></p><p>1) The \: sides \: of \: a \: rhombus \: are \: all \: congruent \: (the \: same \: length.) </p><p>A</p><p>B</p><p>=</p><p>B</p><p>C</p><p>=</p><p>C</p><p>D</p><p>=</p><p>D</p><p>A</p><p>=</p><p>a</p><p> </p><p></p><p>2) Opposite \: angles \: of \: a \: rhombus \: are \: congruent \: (the \: same \: size \: and \: measure.)</p><p>∠</p><p>B</p><p>A</p><p>D</p><p>=</p><p>∠</p><p>B</p><p>C</p><p>D</p><p>=</p><p>y</p><p>,</p><p>and</p><p>∠</p><p>A</p><p>B</p><p>C</p><p>=</p><p>∠</p><p>A</p><p>D</p><p>C</p><p>=</p><p>x</p><p></p><p>3) The \: intersection \: of \: the diagonals \: of \: a \: rhombus form \: 90 \: degree \: (right) \: angles. \: This \: means \: that \: they \: are \: perpendicular.</p><p>∠</p><p>A</p><p>O</p><p>B</p><p>=</p><p>∠</p><p>B</p><p>O</p><p>C</p><p>=</p><p>∠</p><p>C</p><p>O</p><p>D</p><p>=</p><p>∠</p><p>D</p><p>O</p><p>A</p><p>=</p><p>90</p><p>∘</p><p> </p><p></p><p>4) The \: diagonals \: of \: a \: rhombus \: bisect \: each \: other. This \: means \: that \: they \: cut \: each other \: in \: half.</p><p>B</p><p>O</p><p>=</p><p>O</p><p>D</p><p>=</p><p>1</p><p>2</p><p>B</p><p>D</p><p>=</p><p>m</p><p>,</p><p>and</p><p>A</p><p>O</p><p>=</p><p>O</p><p>C</p><p>=</p><p>1</p><p>2</p><p>A</p><p>C</p><p>=</p><p>n</p><p></p><p>5) Adjacent \: sides \: of \: a rhombus \: are \: supplementary. \: This \: means that \: their \: measures \: add up \: to \: 180 \: degrees.</p><p>x</p><p>+</p><p>y</p><p>=</p><p>180</p><p>∘</p><p> </p><p></p><p>Now \: back to \: our \: question.</p><p></p><p>In </p><p>Δ</p><p>B</p><p>O</p><p>C</p><p>,</p><p>B</p><p>C</p><p>2</p><p>=</p><p>B</p><p>O</p><p>2</p><p>+</p><p>O</p><p>C</p><p>2</p><p> </p><p>Since </p><p>B</p><p>O</p><p>=</p><p>1</p><p>2</p><p>B</p><p>D</p><p>,</p><p>and</p><p>O</p><p>C</p><p>=</p><p>1</p><p>2</p><p>A</p><p>C</p><p> </p><p></p><p>⇒</p><p>B</p><p>C</p><p>2</p><p>=</p><p>(</p><p>1</p><p>2</p><p>B</p><p>D</p><p>)</p><p>2</p><p>+</p><p>(</p><p>1</p><p>2</p><p>A</p><p>C</p><p>)</p><p>2</p><p> </p><p></p><p>⇒</p><p>B</p><p>C</p><p>2</p><p>=</p><p>1</p><p>4</p><p>(</p><p>B</p><p>D</p><p>)</p><p>2</p><p>+</p><p>1</p><p>4</p><p>(</p><p>A</p><p>C</p><p>)</p><p>2</p><p> </p><p></p><p>⇒</p><p>B</p><p>C</p><p>2</p><p>=</p><p>1</p><p>4</p><p>(</p><p>B</p><p>D</p><p>2</p><p>+</p><p>A</p><p>C</p><p>2</p><p>)</p><p> </p><p></p><p>⇒</p><p>4</p><p>B</p><p>C</p><p>2</p><p>=</p><p>A</p><p>C</p><p>2</p><p>+</p><p>B</p><p>D</p><p>2</p><p>$

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Answer 2

Answer:

In rhombus ABCD, prove that 4AB2 = AC2 + BD2. Proof: The diagonals of a rhombus bisect each other at right angles.