Draw an equilateral $\triangle BDP$ $\because BD=BP, \angle PBA=60^{\circ}-\angle ABD=\angle DBC,BC=BA$ $\therefore \triangle APB \cong \triangle CDB$ so $AP=CD$ and $\angle PAD$ $=\angle BAP+\angle BAD$ $=\angle BCD+\angle BAD$ $=60^{\circ}-\angle ACD+\angle BAD$ $=30^{\circ}+\angle CAD+\angle BAD$ $=90^{\circ}$ ${|AD|, |PD|, |AP|}$ is necessarily a right-angled triangle. so ${|AD|, |BD|, |CD|}$ is necessarily a right-angled triangle.

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