PROBLEM IX.

A geometric diagram displays a circle with center B. A line segment AD passes through the circle, intersecting it at points E and Z. Line segments AB and BE are drawn to the center B, creating a triangle related to the circle.

Given two lines AB = 56 = b and AD = 92 = d, form a triangle in such a way that AB × ED (the difference of the sides = x) is equal to the square of ZD (that is) the square of the difference of the bases. What is (ZD =) a?

From the nature of the Problem, bx = aa: It will be that ED = x = aa/b, and in every triangle, AB + BE + ED × ED, that is 2AB + ED × ED = AD × ZD, i.e., 2b + aa/b × aa/b = da, and (2bbaa + aaaa) / bb = da, and aaa + 2bba = bbd.

That is aaa + ba = c. In numbers aaa + 6272a = 288512.

Theor. x = (c - ggg - bg) / (3gg + b)

$\qquad g = 4 \qquad b = 62 \qquad c = 288$
$+ c - ggg - bg = - 24$
$3gg + b = 110) - 240 (- 02 = x$

$\qquad 40$
$\quad + 02$
$\qquad g = 38,$
$+ c - ggg - bg = - 4696$
$3gg + b = 1064) - 4696.0 (-, 44 = x$

$\qquad 38,$
$\quad -, 44$
$\qquad g = 37,56$
$+ c - ggg - bg = - 52,225216$
$3gg + b = 10504) - 52,225216 (-, 004971 = x$

$\qquad 37,56$
$\quad -, 004971$
$\qquad a = 37,555029$
$+ c - ggg - bg = -, 0113198757926289389$
$3gg + b = 10503,14) -, 0113198757926289389 (-, 0000010777593 = x$

$\qquad 37,555029$
$\quad -, 0000010777593$
$\qquad a = 37,5550279222407$