An interesting geometric property of the circle

Let $AC$ be a diameter of an arbitrary circle, let $B$ be any point on the circle, thus defining a triangle $ABC$ inscribed in the circle, and let $BD$ be the altitude of this triangle.

Show that the area of the square with side length $BD$ is equal to the area of the rectangle with length $AD$ and width $CD$ ; in other words, show that $AD/BD=BD/DC$ . (*)

(*) For ancient Greek and Arab mathematicians, $BD$ was called the “mean proportional” between $AD$ and $DC$ .

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Solution by alouette · translated by Ancient Tree

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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 $DC$ .

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