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13/07/2021 11:43 am
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Diagonal AC of a parallelogram ABCD bisects ∠A (see Figure). Show that
(i) it bisects ∠C also,
(ii) ABCD is a rhombus.
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13/07/2021 11:45 am
(i) In ΔADC and ΔCBA,
AD = CB (Opposite sides of a parallelogram)
DC = BA (Opposite sides of a parallelogram)
AC = CA (Common Side)
ΔADC ≅ ΔCBA [SSS congruency]
∠ACD = ∠CAB by CPCT
and ∠CAB = ∠CAD (Given)
⇒ ∠ACD = ∠BCA
AC bisects ∠C also.
(ii) ∠ACD = ∠CAD (Proved above)
⇒ AD = CD (Opposite sides of equal angles of a triangle are equal)
Also, AB = BC = CD = DA (Opposite sides of a parallelogram)
ABCD is a rhombus.