Question
$\text{ABCD}$ is a rhombus. Prove that $AB^2 + BC^2 + CD^2 + DA^2= AC^2 + BD^2$
✓
Answer
Let the diagonals $A C$ and $B D$ of rhombus $\text{ABCD}$ intersect at $O$.
Since the diagonals of a rhombus bisect each other at right angles.
$\therefore \angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}$ and $A O=C O, B O=O D$.
Since $\triangle A O B$ is a right triangle right$-$angle at $O$.
$ \therefore AB ^2= OA ^2+ OB ^2$
$A B^2=\left(\frac{1}{2} A C\right)^2+\left(\frac{1}{2} B D\right)^2$
$\Rightarrow 4 AB ^2= AC ^2+ BD ^2 \ldots \text { (i) } $
Similarly, we have
$ 4 B C^2=A C^2+B D^2 \ldots(\text { ii })$
$4 C D^2=A C^2+B D^2 \ldots(\text { iii }) $
and,
$4 A D^2=A C^2+B D^2 \ldots(i v)$
Adding all these results, we get
$ 4\left(A B^2+B C^2+A D^2\right)=4\left(A C^2+B D^2\right)$
$\Rightarrow A B^2+B C^2+A D^2+D A^2=A C^2+B D^2 $
Since the diagonals of a rhombus bisect each other at right angles.
$\therefore \angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}$ and $A O=C O, B O=O D$.
Since $\triangle A O B$ is a right triangle right$-$angle at $O$.
$ \therefore AB ^2= OA ^2+ OB ^2$
$A B^2=\left(\frac{1}{2} A C\right)^2+\left(\frac{1}{2} B D\right)^2$
$\Rightarrow 4 AB ^2= AC ^2+ BD ^2 \ldots \text { (i) } $
Similarly, we have
$ 4 B C^2=A C^2+B D^2 \ldots(\text { ii })$
$4 C D^2=A C^2+B D^2 \ldots(\text { iii }) $
and,
$4 A D^2=A C^2+B D^2 \ldots(i v)$
Adding all these results, we get
$ 4\left(A B^2+B C^2+A D^2\right)=4\left(A C^2+B D^2\right)$
$\Rightarrow A B^2+B C^2+A D^2+D A^2=A C^2+B D^2 $
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