AB and DC are two equal chords of a circle. To prove: AC = BD Construction: Join A to C and B to D. Let AC and BD intersect at point P. Proof: Since ∠BAC and ∠BDC lie on the same arc of the circle, ∴ ∠BAC = ∠BDC That is, ∠BAP = ∠PDC Now, in triangles ∆APB and ∆DPC: AB = DC [Given] ∠APB = vertically opposite ∠DPC And ∠BAP = ∠PDC [Proved earlier] ∴ ∆APB ≅ ∆DPC ∴ AP = DP and PB = PC [Corresponding sides] ∴ AP + PC = DP + PB That is, AC = BD (Proved)