First take reflection, then cyclic

If the right-angled triangle $ABD$ has sides $AB = x$ , $AD = y = 999$ and $BD = \sqrt{x^2 + y^2}$ , then $ABD$ has semiperimeter $s = \tfrac12\big(x + y + \sqrt{x^2+y^2}\big)$ . The inradius $r$ of this triangle is the distance from $A$ to the point of tangency of the incircle with $AD$ (or with $AB$ ), and this is equal to $r \; = \; s - BD \; = \; \tfrac12\big(x + y - \sqrt{x^2+y^2}\big) \;.$ But then the distance $DP$ from $D$ to $P$ , the point of tangency of the incircle with $BD$ , is $DP \; = \; s - x \; = \; \tfrac12\big(y - x + \sqrt{x^2+y^2}\big) \; = \; y - r \;.$ On the other hand $DQ$ , the distance from $D$ to the point of tangency of the incircle of $BDC$ with $BD$ , is equal to the inradius $R$ of the triangle $BDC$ . Thus $PQ \; = \; DP - DQ \; = \; y - r - R$ and hence $r + R \,=\, y - PQ \,=\, AD - PQ \,=\, \boxed{799}$ .