question about right triangle Suppose there is given right triangle($AC=6$) and circle (radius$=5$) which goes through  $A$ and $C$ and meets $AB$ leg at the midpoint. Here is picture

We are asked to find length of leg.in my point of view, we can connect center of circle with $A$ and $C$ point,get isosceles  triangle we know all sides ,find angle at  center, then connect center also to $A$ and $D$  here we know both length (radius) angle will be $180-a$ ($a$ it is which we got by cosine law) calculate $AD$  by the cosine law   and got finally $AB$, can you show me shortest way? or am I correct or wrong? Please help!
 A: Let $S$ be the center of the circle and $M$ be the midpoint of $AC$.
From the right triangle AMS we can get:
$|MS|=\sqrt{5^2-3^2}=4$.
Now we use the right triangle AMD. (This triangle is right since D is the midpoint of AB - have a look at similar triangles ADD' and ACB, where D' is the point of AC such that DD' is perpendicular to AC. You should see that D'=M.)
This right triangle gives us:
$|AD|=\sqrt{3^2+9^2}=3\sqrt{10}$ and $|AB|=2|AD|=6\sqrt{10}$

What is correct English terminology for this: "D' is the point of AC such that DD' is perpendicular to AC"? If I used word by word translation from my language, it would be "D' is the foot of the perpendicular from the point D to the line AC".

If I try to follow your suggestions and compute the angles then I get:
$\sin\alpha=\frac35$ and $\cos\alpha=\frac45$ ($\alpha$ denotes the angle ASM)
Now I digress a little from your suggestion.
$\frac{|AB|}4=5\cos\frac\alpha2=5\sqrt{\frac{1+\cos\alpha}2}=3\sqrt{\frac52}$
$|AB|=4.3\sqrt{\frac52}=6\sqrt{10}$.
A: $$\left\vert AC\right\vert ^{2}+\left\vert BC\right\vert ^{2}=\left\vert
AB\right\vert ^{2}=4\left\vert AD\right\vert ^{2}$$
$$6^{2}+\left\vert BC\right\vert ^{2}=4\left\vert AD\right\vert ^{2}$$
$$h^{2}=5^{2}-3^{2}=16,$$
where $h$ is the distance from the center to $AC$. Hence $h=4$ and
$$|AD|^2=\left( 5+h\right) ^{2}+3^{2}=9^{2}+3^{2}=90.$$
Thus
$$\left\vert AD\right\vert =3\sqrt{10}.$$
And $|BC|$ is such that
$$6^{2}+\left\vert BC\right\vert ^{2}=4\cdot 90,$$
$$\left\vert BC\right\vert =18.$$
