Question

prove that the diagonals of a rhombus bisect each other at right angles​

Answer 1

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$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

prove that the diagonals of a rhombus bisect each other at right angles

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${\blue{\sf\underline{Given}}}$

A rhombus ABCD whose diagonals AC and BD intersect at a point O.

${\green{\sf\underline{To\:Prove}}}$

(i) OA = OC and OB = OD

(ii) ∠BOC = ∠DOC = ∠AOD = ∠AOB = 90°

${\blue{\sf\underline{Proof}}}$

(i) Clearly, ABCD is a ||gm, in which AB = BC = Cd = DA

Also, we have that the diagonals of a ||gm bisect each other.

∴OA = OC and OB = OD.

(ii) Now, in ∆BOC and DOC,we have

OB = OD, BC = DC and OC = OC (common)

∴ ∆BOC ≅ ∆DOC.

∴ ∠BOC = ∠DOC (c.p.c.t.).

But, ∠BOC + ∠DOC = 180° (linear pair)

∴ ∠BOC = ∠DOC = 90°.

Similarly,∠AOB = ∠AOD = 90°.

Hence, the diagonals of a rhombus bisect each other at right angles

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Answer 2

Answer:

$Let ABCD is a rhombus.</p><p></p><p>⇒  AB=BC=CD=DA                     [ Adjacent sides are eqaul in rhombus ]</p><p></p><p>In △AOD and △COD</p><p></p><p>⇒  OA=OC          [ Diagonals of rhombus bisect each other ]</p><p></p><p>⇒  OD=OD         [ Common side ]</p><p></p><p>⇒  AD=CD         </p><p></p><p>∴  △AOD≅△COD              [ By SSS congruence rule ]</p><p></p><p>⇒  ∠AOD=∠COD                [ CPCT ]</p><p></p><p>⇒  ∠AOD+∠COD=180o             [ Linear pair ]</p><p></p><p>⇒  2∠AOD=180o.</p><p></p><p>∴  ∠AOD=90o.</p><p></p><p>Hence, the diagonals of a rhombus bisect each other at right angle.</p><p></p><p>$