Question

If a diagonal of a parallelogram bisects one of the angles of the parallelogram prove, that it also bisects the angle opposite it and that the two diagonals are perpendicular to each other also prove that it is a rhombus.​

Answer 1

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$\huge{\underline{\underline{\sf{\orange{</p><p>Question:-}}}}}$

If a diagonal of a parallelogram bisects one of the angles of the parallelogram prove, that it also bisects the angle opposite it and that the two diagonals are perpendicular to each other also prove that it is a rhombus.

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$\huge{\underline{\underline{\sf{\orange{</p><p>Solution:-}}}}}$

${\blue{\sf\underline{Given}}}$

A ||gm ABCD whose diagonals AC and BC intersect at O. Also, AC bisects ∠A , i.e. , ∠1 = ∠2.

${\blue{\sf\underline{To\:prove}}}$

(i) AC bisects ∠C , i.e. , ∠3 = ∠4.

(ii) AC ⊥ BD.

(iii) ABCD is a rhombus.

${\blue{\sf\underline{Proof}}}$

In ||gm ABCD,we have

AB || DC and CA cuts them

∴∠1 = ∠3 (alt.interior ∠s)

And AD || BC and cuts them

∴∠2 = ∠4 (alt.interior ∠s)

Now,∠1 = ∠2 ⇒ ∠3 = ∠4.

This shows that AC bisects ∠C also.

Now,∠A = ∠C (opp.∠s of ||gm).

$⇒ \frac{1}{2} \angle \: A = \frac{1}{2} \angle \: C ⇒ \angle \: 2 = \angle \: 3 ⇒ AD = DC$

But, AB = DC and DC = BC .

∴AB = BC = DC = AD.

So, ABCD is a rhombus.

But the diagonals of a rhombus bisect each other at right angles.

∴AC ⊥ BD.

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Answer 2

To prove

(i) AC bisects ∠C , i.e. , ∠3 = ∠4.

(ii) AC ⊥ BD.

(iii) ABCD is a rhombus.

Proof

In ||gm ABCD,we have

AB || DC and CA cuts them

∴∠1 = ∠3 (alt.interior ∠s)

And AD || BC and cuts them

∴∠2 = ∠4 (alt.interior ∠s)

Now,∠1 = ∠2 ⇒ ∠3 = ∠4.

This shows that AC bisects ∠C also.

Now,∠A = ∠C (opp.∠s of ||gm).

\ ⇒ 1/2 < A = 1/2 < C

⇒ < 2 = < 3

⇒ AD = DC

⇒ 21 ∠ A= 21

∠C⇒∠2=∠3⇒AD=DC

But, AB = DC and DC = BC .

∴AB = BC = DC = AD.

So, ABCD is a rhombus.

But the diagonals of a rhombus bisect each other at right angles.

∴AC ⊥ BD.