Math, asked by tchourasiya2002, 1 year ago

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


tchourasiya2002: Guys answer it fast

Answers

Answered by Suchitrasihag1
19
hope this will help you
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tchourasiya2002: But s.s.a is not a congruency rule
Suchitrasihag1: it is a congruency rule
Answered by TIGER1407
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.

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