3. In an equilateral triangle $\mathrm{ABC}$, the height $\mathrm{BH}$ is drawn. On the line $\mathrm{BH}$, a point $\mathrm{D}$ is marked such that $\mathrm{BD}=\mathrm{AB}$. Find $\angle \mathrm{CAD}$.
Answer: $15^{\circ}$ or $75^{\circ}$.

Solution. Obviously, $\triangle \mathrm{ADH}=\Delta \mathrm{CDH}$ (right-angled, by two legs), hence $\mathrm{AD}=\mathrm{CD}$. Therefore, $\triangle \mathrm{BDA}=\triangle \mathrm{BDC}$ (by three sides) $\Rightarrow \angle \mathrm{DAB}=\angle \mathrm{DCB}, \angle \mathrm{BDA}=\angle \mathrm{BDC}$.

Moreover, $\triangle \mathrm{ABD}$ is isosceles $(\mathrm{BD}=\mathrm{AB}) \Rightarrow \angle \mathrm{BDA}=\angle \mathrm{BAD}$, that is, $\angle \mathrm{BDC}=\angle \mathrm{BDA}=\angle \mathrm{BAD}=\angle \mathrm{DCB}=\alpha$.

Consider $\triangle \mathrm{ADC}$ and the sum of its angles.

Next, we need to consider two cases for the position of point $\mathrm{D}$: on the same side as point B relative to AC and on opposite sides. In the first case,

$\angle \mathrm{CDA}+\angle \mathrm{DAC}+\angle \mathrm{DCA}=\angle \mathrm{CDB}+\angle \mathrm{BDA}+\angle \mathrm{DAB}+\angle \mathrm{BAC}+\angle \mathrm{DCB}+\angle \mathrm{CBA}=$ $=\alpha+\alpha+\alpha+60^{\circ}+\alpha+60^{\circ}=4 \cdot \alpha+120^{\circ}=180^{\circ}$ (sum of angles in a triangle), from which $\alpha=60^{\circ} / 4=15^{\circ} . \angle \mathrm{CAD}=\angle \mathrm{CAB}+\angle \mathrm{BAD}=60^{\circ}+\alpha=75^{\circ}$.

In the second case,

$\angle \mathrm{CDA}+\angle \mathrm{DAC}+\angle \mathrm{DCA}=\angle \mathrm{CDB}+\angle \mathrm{BDA}+\angle \mathrm{DAB}-\angle \mathrm{BAC}+\angle \mathrm{DCB}-\angle \mathrm{CBA}=$ $=\alpha+\alpha+\alpha-60^{\circ}+\alpha-60^{\circ}=4 \cdot \alpha-120^{\circ}=180^{\circ}$ (sum of angles in a triangle), from which $\alpha=300^{\circ} / 4=75^{\circ} . \angle \mathrm{CAD}=\angle \mathrm{BAD}-\angle \mathrm{BAC}=\alpha-60^{\circ}=15^{\circ}$.

Comment. Only the answer - 0 points. Only one case of the position of point $\mathrm{D}$ is considered (correspondingly, only one value is obtained) - 4 points.