Q-17 plz answer..its very urgent ...tomorrow is my maths exam....
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Hey,
Given: ABCD is a parallelogram
To prove: ∠A = ∠C and ∠B = ∠D
Proof: AB||CD and AC is the transversal
Hence ∠1 = ∠4 → (1)
ΔABC ≅ ΔADC (SSS Congruence rule)
Hence ∠3 = ∠4 → (2) and ∠1 = ∠2 → (3)
Consider, ∠A = ∠3 + ∠4 = ∠4 + ∠4 [From (2)]
= 2∠4
= 2∠1
= ∠1 + ∠2
= ∠C
Therefore, ∠A = ∠C Similarly, we can prove ∠B = ∠D.
HOPE IT HELPS YOU:-))
Given: ABCD is a parallelogram
To prove: ∠A = ∠C and ∠B = ∠D
Proof: AB||CD and AC is the transversal
Hence ∠1 = ∠4 → (1)
ΔABC ≅ ΔADC (SSS Congruence rule)
Hence ∠3 = ∠4 → (2) and ∠1 = ∠2 → (3)
Consider, ∠A = ∠3 + ∠4 = ∠4 + ∠4 [From (2)]
= 2∠4
= 2∠1
= ∠1 + ∠2
= ∠C
Therefore, ∠A = ∠C Similarly, we can prove ∠B = ∠D.
HOPE IT HELPS YOU:-))
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hey mate here is your answer ...
please mark as a brainliest...
all the best for the exam.....
please mark as a brainliest...
all the best for the exam.....
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