Question

2] Prove that -
Diagonals of a rhombus bisect its opposite angle.
Given : ABCD is a rhombus.
seg AC is a diagonal.
Q
To prove : <BAC DAC
< BCA = DCA
Proof: In AABC & A ADC
side AB
side BC
side AC = side AC
A ABC = A ADC
< BAC =< DAC
c.9.c.t
< BCA =<DCA​

Answer 1

Step-by-step explanation:

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

[ Linear pair ]

⇒ 2∠AOD=180

o ∠AOD=90

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