Given:
$$AD$$ is the bisector of $$\angle A$$
$$\Rightarrow\angle DAB=\angle DAC$$ ...$$(1)$$
$$AD\bot BC$$
$$\Rightarrow\angle BDA=\angle CDA={90}^{o}$$
To prove:
$$\triangle ABC$$ is isosceles.
Proof:
In $$\triangle DAB$$ and $$\triangle DAC$$,
$$\angle BDA=\angle CDA={90}^{o}$$
$$DA=DA$$ [common]
$$\angle DAB=\angle DAC$$ [from $$(1)$$]
$$\therefore\triangle DAB\cong \triangle DAC$$ [By ASA congruence property]
$$\Rightarrow AB=AC$$
Hence, $$\triangle ABC$$ is isosceles