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