In $$\triangle$$ ABC, $$AD\perp BC$$ at D and AE is the bisector of $$\angle BAC$$, $$\angle$$B = $$72^\circ$$ and $$\angle$$C = $$26^\circ$$, then what is the measure of $$\angle$$DAE ?

As per the question,
It is given that, $$AD\perp BC$$
AE is the bisector of $$\angle BAC$$
$$\angle$$B = $$72^\circ$$
$$\angle$$C = $$26^\circ$$
Now, As per the diagram,

In, $$\triangle ABC$$, $$\angle A + \angle B +\angle C=180$$

$$= \angle A=180 -(\angle B +\angle C)$$

$$= \angle A=180 -(72+26)$$

$$= \angle A=82$$

as per the question, $$AE$$ is the bisector of $$\angle A$$,

So,$$\angle BAE=\angle CAE = 41^\circ$$

Now, In $$\triangle ADC$$

$$\angle DAC +\angle D +\angle C=180$$

$$\angle DAC =180-\angle D -\angle C$$

$$\angle DAC =180-90-26=64^\circ$$

Hence, $$\angle DAE = \angle DAC - \angle CAE = 64^\circ - 41^\circ =23^\circ $$