Diagonal AC of a parallelogram ABCD bisects ∠A (see the given figure). Show that
(i) It bisects ∠C also,
(ii) ABCD is a rhombus.
(i) ABCD is a parallelogram.
∴ ∠DAC = ∠BCA (Alternate interior angles) ... (1)
And, ∠BAC = ∠DCA (Alternate interior angles) ... (2)
However, it is given that AC bisects ∠A.
∴ ∠DAC = ∠BAC ... (3)
From equations (1), (2), and (3), we obtain
∠DAC = ∠BCA = ∠BAC = ∠DCA ... (4)
⇒ ∠DCA = ∠BCA
Hence, AC bisects ∠C.
(ii)From equation (4), we obtain
∠DAC = ∠DCA
∴ DA = DC (Side opposite to equal angles are equal)
However, DA = BC and AB = CD (Opposite sides of a parallelogram)
∴ AB = BC = CD = DA
Hence, ABCD is a rhombus.
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