$A$ circle has a radius of $\sqrt{2} \text{ cm}$. It is divided into two segments by a chord of length $2 \text{ cm}$. Prove that the angle subtended by the chord at a point in the major segment is $45^{\circ}$.

(N/A) Let the circle have center $O$ and radius $r = \sqrt{2} \text{ cm}$. Let $BC$ be the chord of length $2 \text{ cm}$.
Join $OB$ and $OC$. In $\triangle OBC$,we have $OB = OC = \sqrt{2} \text{ cm}$ and $BC = 2 \text{ cm}$.
Check if $\triangle OBC$ is a right-angled triangle:
$OB^2 + OC^2 = (\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4$.
$BC^2 = (2)^2 = 4$.
Since $OB^2 + OC^2 = BC^2$,by the converse of the Pythagoras theorem,$\angle BOC = 90^{\circ}$.
The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,$\angle BAC = \frac{1}{2} \angle BOC = \frac{1}{2} \times 90^{\circ} = 45^{\circ}$.
Hence,it is proved that the angle subtended by the chord in the major segment is $45^{\circ}$.