Question

The diagonals of a parallelogram bisects one of its angles. Show that its a rhombus.

Answer 1

take two adjacent triangles in which diagonal bisecting one of it's diagonals

as diagonals bisect each other

so 2 corresponding sides are equal
Also angles are equal as diagonal bisect one of it's angle
so SAS SIMILARITY

THEREFORE third side becomes equal

that is adjacent sides of parallelogram are equal
that means all four sides are equal

So it's rhombus

Answer 2

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.