Question

$\pink{ \underline{ \large{ \frak{Q uestion : }}}}$
Prove that each diagonal of a rhombus bisect each other at 90°


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Answer 1

Answer:

❥˙Therefore, first we should show that the triangles made at opposite sides by the diagonals and sides are congruent and use it to show that the diagonals bisect each other. Then, we shall again use the property of congruence to prove that the angle between the diagonals is 90∘. ... Thus, the diagonals bisect each other.✧*。

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Answer 2

$\pink{ \underline{ \large\huge \underline \color{red}{ \frak{Q uestion : }}}}$

• Prove that each diagonal of a rhombus bisect each other at 90°

$\pink{ \underline{ \large\huge \underline \color{red}{ \frak{Answer :}}}}$

To Prove :

• That each diagonal of a rhombus bisect each other at 90.

Solution :

➥ Let the rhombus be ABCD.

⟹ AB = DA [ Adjacent sides are equal ]

In △AOD and △COD

⟹ OA=OC [ diagonals bisects at each other ]

⟹ OD=OD [ common point ]

⟹ AD=CD

➥ Therefore : △AOD =△COD

⟹ ∠AOD = ∠COD

⟹ ∠AOD + ∠COD = 180°

⟹ 2∠AOD = 180°

⟹ ∠AOD = 180÷2

$\underline \color{cyan} \: = \: 90°\color{silver}☆$

↬ Hence, proved that the diagonals of a rhombus bisect each other at right angle.