Question
In the given figure, $ABCD$ is a Rhombus. Then,
✓
Answer
$ABCD$ is a rhombus. $AB = BC = CD = DA$
In Rhombus, diagonals bisect each other at right angles.
So, $AO= CO$ and $BO = DO$
In triangle $AOB \times AO^2 + BO^2 = AB^2 ($Pythagoras theorem$)$
$\Big(\frac{1}{2}\text{AC}\Big)^2 + \Big(\frac{1}{2}\text{BD}\Big)^2 = \text{AB}^2$
$\frac{\text{AC}^2}{4} + \frac{\text{BD}^2}{4} = \text{AB}^2$
$= AC^2 + BD^2 = 4AB^2$
In Rhombus, diagonals bisect each other at right angles.
So, $AO= CO$ and $BO = DO$
In triangle $AOB \times AO^2 + BO^2 = AB^2 ($Pythagoras theorem$)$
$\Big(\frac{1}{2}\text{AC}\Big)^2 + \Big(\frac{1}{2}\text{BD}\Big)^2 = \text{AB}^2$
$\frac{\text{AC}^2}{4} + \frac{\text{BD}^2}{4} = \text{AB}^2$
$= AC^2 + BD^2 = 4AB^2$
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