In fig 3.39 chord AB ≅ chord CD, Prove that, arc AC ≅ arc BD
Figure: Chords AB and CD intersect inside the circle. Going round the circle the points are in the order A, C, B, D (arcs ACB and CBD are minor arcs).
Answer: Congruent chords give arc ACB ≅ arc CBD; subtracting the common arc CB gives arc AC ≅ arc BD.
Chord AB ≅ chord CD (given).
\(\therefore\) arc ACB ≅ arc CBD (arcs corresponding to congruent chords).
\[m(\text{arc ACB}) = m(\text{arc CBD})\] \[m(\text{arc AC}) + m(\text{arc CB}) = m(\text{arc CB}) + m(\text{arc BD})\] \[m(\text{arc AC}) = m(\text{arc BD})\]Arcs of the same circle with equal measures are congruent, so arc AC ≅ arc BD.
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