Math, asked by mamtakumariaf, 6 months ago

Show that a parallelogram is a rhombus if its diagonals bisect opposite angles.

Answers

Answered by hemayet
0

Answer:

To prove that AC ⊥ BD. DAB. 'The diagonals of a rhombus bisect each other at right angles. ... We have proved that the opposite sides of a parallelogram are equal, so if two adjacent sides are equal, then all four sides are equal and it is a rhombus

Step-by-step explanation:

if two adjacent sides are equal then all the four sides are equal , it is called rhombus

Answered by TIGER1407
0

Answer:

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Step-by-step explanation:

Given: ABCD is a parallelogram and diagonal AC bisects ∠A.

To prove: Diagonal AC bisects ∠A ∠1 = ∠2

Now, AB || CD and AC is a transversal.

∠2 = ∠3 (alternate interior angle) Again AD || BC and AC is a transversal.

∠1 = ∠4 (alternate interior angles)

Now, ∠A = ∠C (opposite angles of a parallelogram)

⇒ 1/2∠A = 1/2 ∠C

⇒ ∠1 = ∠3 ⇒ AD = CD (side opposite to equal angles)

AB = CD and AD = BC AB = BC = CD = AD ⇒ ABCD is a rhombus.

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