Question

of a parallelogram bisects one of its angles. Show that it is a rhombus.​

Answer 1

Step-by-step explanation:

The statement , diagonal of a parallelogram bisects one of its angle, then it is a rhombus it is correct.

Answer 2

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) ⇒ 12 1 2 ∠A = 12 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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