The midpoint of a chord of length 2 a 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 a 2 + d 2 d .I know I have to find similar triangles, I cannot see them...

Question

The midpoint of a chord of length   2  a 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                     a        2            +              d        2              d   .I know I have to find similar triangles, I cannot see them...

Gilbert Petty · Accepted Answer

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   C  D is a diameter of the circle, and   H the intersection point of   C  D with the chord   A  B.In the right triangle   C  A  D, the altitude   A  H cuts out two segments   C  H and   H  D on the hypotenuse   C  D and we know the altitude is the geometric mean of these two segments, in other words  A      H    2    =      a    2    =  C  H  ⋅  H  C  =  d  (  C  D  −  d  )  ,    so     d    C  D  =      a    2    +      d    2    ,whence the formula.

he 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 (a^2+d^2)/d .

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