Question

6. Diagonal AC of a parallelogram ABCD bisects
2 A(see Fig. 8.19). Show that
mit bisects 2 C also,
(ii) ABCD is a rhombus.​

Answer 1

Step-by-step explanation:

(i) Here, ABCD is a parallelogram and diagonal AC bisects ∠A.

∴ ∠DAC=∠BAC ---- ( 1 )

Now,

AB∥DC and AC as traversal,

∴ ∠BAC=∠DCA [ Alternate angles ] --- ( 2 )

AD∥BC and AAC as traversal,

∴ ∠DAC=∠BCA [ Alternate angles ] --- ( 3 )

From ( 1 ), ( 2 ) and ( 3 )

∠DAC=∠BAC=∠DCA=∠BCA

∴ ∠DCA=∠BCA

Hence, AC bisects ∠C.

(ii) In △ABC,

⇒ ∠BAC=∠BCA [ Proved in above ]

⇒ BC=AB [ Sides opposite to equal angles are equal ] --- ( 1 )

⇒ Also, AB=CD and AD=BC [ Opposite sides of parallelogram are equal ] ---- ( 2 )

From ( 1 ) and ( 2 ),

⇒ AB=BC=CD=DA

Hence, ABCD is a rhombus.