Question

a diagonal of a parallogram bisects one of its angles .show that it is rhombus​

Answer 1

Given:

✰ A diagonal of a parallogram bisects one of its angles.

To proof:

✠ It is a rhombus.

Solution:

Let ABCD is a parallogram and diagonal AC bisects one of its angles i.e, ∠A

⤳ ∠1 = ∠2

AD is parallel to BC ( AD || BC ) and AC is a transversal

⤳ ∠1 = ∠4 ( alternate interior angles )

AB is parallel to CD ( AB || CD ) and AC is a transversal

⤳ ∠2 = ∠3 ( alternate interior angles )

⤳∠A = ∠C ( opposite angles of parallelogram are equal )A

Also, 1/2 ∠A = 1/2 ∠C

∴ ∠1 = ∠3 ( They are nothing but 1/2 ∠A and 1/2 ∠C )

⤳ AD = CD ( Side opposite to the equal angles are equal. )

⤳ AB = CD

Also, AD = BC

( ∵ Opposite sides of parallelogram are equal )

∴ AB = BC = CD = AD

Therefore, it is a rhombus.

Hence Proved!!

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