The value of angle BAD in trapezoid ABCD, where AB is parallel to CD, BC is perpendicular to AC, and BC equals AC, is determined to be 45 degrees. This conclusion arises from the properties of isosceles right triangles formed by the given conditions. Specifically, since BC and AC are equal and form a right angle, triangle ABC is an isosceles right triangle, leading to the conclusion that angle BAD must also be 45 degrees.
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Albert said:A trapezoid ABCD given :
(1)$\overline{AB} // \overline{CD}$
(2)$\overline{BC}\ \perp \overline{AC}$ and $\overline{BC}=\overline{AC}$
(3)$\overline{AB}=\overline{BD}$
find the value of $\angle BAD$
very good solution!greg1313 said:The height of trapezoid $$ABCD$$ is one-half of its base. This implies that $$\angle{ABD}=30^\circ$$. As $$\triangle{ABD}$$ is isosceles, $$\angle{BAD}=75^\circ$$.