In Fig. 6.26, AB = AD and ∠BAC = ∠DAC. Then

(i) Δ ADC ≅ ΔABC
(ii) BC = DC
(iii) ∠BCA = ∠DCA
(iv) Line segment AC bisects ∠BAD and ∠BCD
Given AB = AD and ∠BAC = ∠DAC with common side AC, ΔADC ≅ ΔABC by SAS criterion. Thus, BC = DC and corresponding angles are equal, so AC bisects ∠BAD and ∠BCD.