Question

Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.

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Answer 1

Step-by-step explanation:

Answer

(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.