diagonal AC of parallelogram ABCD bisect angle A. show that the parallelogram ABCD is a rhombus
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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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