6. Diagonal AC of a parallelogram ABCD bisects ∠A (see Fig. 8.19). Show that (i) it bisects ∠C also, (ii) ABCD is a rhombus.
Solution: (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] Thus, ∠ACD = ∠CAB by CPCT and ∠CAB = ∠CAD (Given) ⇒ ∠ACD = ∠BCA Thus, 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) Thus, ABCD is a rhombus.
