Folkpor

In an isosceles trapezoid $ABCD$, the bases $AD$ and $BC$ are 12 and 6 respectively, and the height is 4. Compare the angles $BAC$ and $CAD$.
Since $A D \| B C$, then $\angle C A D=\angle B C A$ (see the figure). Let $B H$ be the height of the trapezoid. Then $A H=1 / 2(A D-B C)=3$; $B H=4$, and from the right triangle $A B H$ we find that $A B=5$.

Thus, in triangle $A B C \quad B C>A B$, so $\angle B A C>\angle B C A$ (the larger angle is opposite the longer side of the triangle). Therefore, $\angle B A C>\angle C A D$.

![](https://cdn.mathpix.com/cropped/2024_05_06_7041536e9c9bacd5befeg-55.jpg?height=232&width=546&top_left_y=1965&top_left_x=756)

## Answer

$\angle B A C>\angle C A D$.