$\angle BAC = 90^{\circ}$

$A$, $B$ and $C$ are all equidistant from $D$ and therefore lie on a circle whose centre is $D$. $BC$ is a diameter of the circle and $\angle BAC$ is therefore the angle subtended by a diameter at a point on the circumference (the angle in the semicircle).

(Alternatively: let $\angle ACD = x$ and show that $\angle DAC = x,\, \angle ADB = 2x$ and $\angle DAB = 1/2 (180^{\circ}) - 2x = 90^{\circ} - x$. Hence $\angle BAC = x + 90^{\circ} - x = 90^{\circ}$.)

*This problem is taken from the UKMT Mathematical Challenges.*

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