Question

diagonal AC of a parallelogram ABCD bisect angle A.show that (a)it bisect angle c also,(b)ABCD is rhombus

please answer this question
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Answer 1

➙Given:-

Diagonal AC of a parallelogram ABCD bisects ∠A

Show that:

(i) It bisects ∠C Also,

(ii) ABCD is a Rhombus.

➙ Solution :-

(i) Here, ABCD is a parallelogram

Diagonal AC bisects ∠A.

∴ ∠DAC = ∠BACㅤㅤㅤ( 1 )

ㅤㅤㅤㅤ

Now :-

AB ∥ DC and AC as traversal.

↦∠BAC = ∠DCA [Alternate angles]ㅤ( 2 )

AD ∥ BC and AAC as traversal.

ㅤㅤㅤㅤ

↦∠DAC = ∠BCA [Alternate angles]ㅤ( 3 )

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

∠DAC = ∠BAC = ∠DCA = ∠BCA

∴ ∠DCA = ∠BCA

Hence, AC bisects ∠C.

ㅤㅤㅤㅤ

(ii) In △ABC,

↬↬ ∠BAC = ∠BCAㅤ [Proved in above]

➙ BC = AB

[ Sides opposite to equal angles are equal ] (1)

ㅤㅤㅤㅤ

➙ 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.

Answer 2

Answer:

Given:

ABCD is a parallelogram.

Diagonal :- AC

To Proof:

It bisects ∠C also,

ABCD is a rhombus.

Solution:

(i) ABCD is a parallelogram.

→∠DAC = ∠BCA (Alternate interior angles) … (1)

→ ∠BAC = ∠DCA (Alternate interior angles)..(2)

However, it is given that AC bisects ∠A.

∴ ∠DAC = ∠BAC … (3)

From equations (1), (2), and (3), we obtain

⇒ ∠DAC = ∠BCA = ∠BAC = ∠DCA … (4)

⇒ ∠DCA = ∠BCA

Hence, AC bisects ∠C.

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(ii)From equation (4), we obtain

→ ∠DAC = ∠DCA

→ DA = DC (Side opposite to equal angles are equal)

However,

DA = BC and AB = CD (Opposite sides of a parallelogram)

∴ AB = BC = CD = DA

Hence, ABCD is a rhombus.

Hence,

AC bisects ∠C.

ABCD is a rhombus.

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