Find diameter of a circle

Question

The midpoint of a chord of length $2a$ is at a distance $d$ from the midpoint of the minor arc it cuts out from the circle. Show that the diameter of the circle is $\frac{a^2+d^2}{d}$ .

I know I have to find similar triangles, I cannot see them...

Answer 1

$r = \sqrt{r^2 - a^2} + d$

$r - d = \sqrt{r^2 - a^2}$

$(r - d)^2 = r^2 - a^2$

$r^2 - 2rd + d^2 = r^2 - a^2$

$-2rd = -a^2 - d^2$

$r = \frac{a^2 + d^2}{2d}$

So $D = \frac{a^2 + d^2}{d}$

Answer 2

If the chord is $AB$ with midpoint $M$, the midpoint of the arc is $P$ and the point diametral opposite to $P$ is $Q$, then the triangles $AMP$ and $QAP$ are similar.

Answer 3

Let $A$, $B$ be the end points of the chord (and of the chords cut out on the circle), $C$ the mid point of the minor arc, $D$ the midpoint of the major arc, so that CD is a diameter of the circle, and $H$ the intersection point of $CD$ with the chord $AB$.

In the right triangle $CAD$, the altitude $AH$ cuts out two segments $CH$ and $HD$ on the hypotenuse $CD$ and we know the altitude is the geometric mean of these two segments, in other words $$AH^2=a^2=CH\cdot HC=d(CD-d),\enspace\text{so }\;d\, CD=a^2+d^2,$$ whence the formula.