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16 Analytical Geometry Page 31

(Exm 3) Circle with radius R is inscribed in triangle abc

Let P = The perimeter of the triangle
[Let] R = [The] radius of the inscribed circle
[Let] x = [The] altitude of triangle
[Let] y = [The] base [of triangle]

Then, since the area of a triangle is equal to its perimeter
into half the radius of the inscribed circle, (Book B. Prop IX)
we have this equation , or xy = Pa, [never] y = Pr/x

But P = x + y + ; or ; and squaring
x² + y² = P² - 2Px + 2xy - 2Py + x² + y²; or 2 Px - 2xy = P² - 2Py; or
(2p - 2y)x = P² - 2Py: hence x = . Now substituting Pr/x
for y; and we have
. [Or] we have
2Px² - 2Prx = P²x - 2P²r or P(2x² - 2rx - Px) = -2Pr)
or or
or
or