Biology, asked by spsujal, 10 days ago

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

Answers

Answered by mukeshkumarrai70152
1

Answer:

PLEASE MARK ME AS BRAINLIST

Explanation:

We can use alternate interior angles property to show that the diagonal AC bisects ∠C and by showing all sides are equal, it can be proved ABCD is a rhombus. i) ABCD is a parallelogram. However, it is given that AC bisects ∠A. Hence, AC bisects ∠C.

Attachments:
Answered by jankal
2

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.

Similar questions