ADE is an isosceles triangle, so \( \angle ADE = \angle AED \). BDE and CDE are right-angled triangles, so \( \angle BDE = \angle CDE = 90^\circ \).<br><br>
Use the given angles in the figure and angle sum of a triangle (\(180^\circ\)) to find each unknown angle step by step.<br><br>
\( \angle BAC \) is found by adding the relevant angles along the straight line or within the triangles.<br><br>
Final Answer: 140&deg;