Question

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

Answer 1

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.