An interesting geometric property of the circle

The triangle $ABC$, whose side $AC$ is a diameter of the circle, is therefore inscribed in the circle. Hence, triangle $ABC$ is right-angled at $B$.

As a result, the altitude $BD$ divides triangle $ABC$ into two smaller triangles, $ABD$ and $BDC$, which are similar (that is, they are scaled versions of one another and have the same shape).

Two similar triangles have equal corresponding angles. In particular, angle $BAD$ is equal to angle $CBD$, and angle $ABD$ is equal to angle $BCD$.

One property of similar triangles is the equality of the ratios of corresponding sides. Let us compute these ratios. For triangle $ABD$, the ratio is $AD/BD$. For triangle $BDC$, it is $BD/DC$.

Thus, we indeed have:
$AD/BD=BD/DC$.

Another way of writing this relation,
$$BD\times BD=AD\times DC$$shows that the area of the square with side length $BD$ is equal to the area of the rectangle with length $AD$ and width $DC$.