Diagonal AC of a parallelogram ABCD bisects angle A.Show that 1.it bisects angle C also. 2.ABCD is a rhombus.
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Step-by-step explanation:
We have a parallelogram ABCD in which diagonal AC bisects ∠A. ⇒ ∠DAC = ∠BAC
(i) To prove that AC bisects ∠C.
∵ABCD is a parallelogram.
∴AB || DC and AC is a transversal.
∴∠l = ∠3 [Alternate interior angles] ...(1) Also, BC || AD and AC is a transversal.
∴∠2 = ∠4 [Alternate interior angles] ...(2) But AC bisects ∠A. [Given]
∴∠1 = ∠2 ...(3)
From (1), (2) and (3), we have
∠3 = ∠4
⇒AC bisects ∠C.
(ii) To prove ABCD is a rhombus.
In ΔABC, we have ∠1 = ∠4 [∵∠1 = ∠2 = ∠4] ⇒ BC = AB Sides opposite to equal angles
are equal] ...(4) Similarly, AD = DC ...(5) But ABCD is a parallelogram [Given] AB = DC [Opposite sides of a parallelogram] ...(6) From (4), (5) and (6), we have
AB = BC = CD = DA
Thus, ABCD is a rhombus.
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