Square ABCD of side a and N on AB. Find radius of congruent incircles of ACN and BCN

Question

ABCD is a square with side a and diagonal AC. The incircles, with radius r, of the triangles ACN and BCN are congruent, with N in AB. What is the radius r in terms of a?

Answer 1

Let $BN = x$. Then, $AN = a-x$ and $CN = \sqrt{x^2+a^2}$. Establish the triangle areas below

$$Area_{NBC} =\frac12 NB \cdot BC = \frac12r\cdot (NB+BC + NC)$$ $$Area_{NAC} =\frac12 NA \cdot BC = \frac12r\cdot (NA+AC + NC)$$

or

$$xa= r(x+a+\sqrt{x^2+a^2})$$ $$(x-a)a= r(a-x+\sqrt2 a+\sqrt{x^2+a^2})$$

Solve to obtain $x=\sqrt{\frac{\sqrt2-1}2}a$ and the radius

$$r = \frac12(1-\sqrt{\sqrt2-1})a$$