If $d_{1}$ and $d_{2}$ $(d_{2} > d_{1})$ are the diameters of two concentric circles and $c$ is the length of a chord of the outer circle which is tangent to the inner circle,prove that $d_{2}^{2} = c^{2} + d_{1}^{2}$.

(N/A) Let $O$ be the common center of the two concentric circles. Let $AB$ be a chord of the outer circle which touches the inner circle at point $C$.
Since the radius is perpendicular to the tangent at the point of contact,$OC \perp AB$.
In the right-angled triangle $\Delta OCB$,by Pythagoras theorem:
$OC^{2} + CB^{2} = OB^{2}$
Here,$OC$ is the radius of the inner circle,so $OC = \frac{d_{1}}{2}$.
$OB$ is the radius of the outer circle,so $OB = \frac{d_{2}}{2}$.
Since the perpendicular from the center to a chord bisects the chord,$CB = \frac{c}{2}$.
Substituting these values in the equation:
$(\frac{d_{1}}{2})^{2} + (\frac{c}{2})^{2} = (\frac{d_{2}}{2})^{2}$
$\frac{d_{1}^{2}}{4} + \frac{c^{2}}{4} = \frac{d_{2}^{2}}{4}$
Multiplying both sides by $4$,we get:
$d_{1}^{2} + c^{2} = d_{2}^{2}$
Hence,$d_{2}^{2} = c^{2} + d_{1}^{2}$.