India competition problems

Let $E$ and $F$ be the incentres of triangles $ABD$ and $CBD$ respectively. Let the incircles of triangles $ABD$ and $CBD$ touch $AC$ in $P$ and $Q$ respectively. If $\angle BDA = \theta$, we see that

$$r' = PD \tan(\theta/2) = QD \cot(\theta/2).$$
Hence
$$PQ = PD + QD = r' \left( \cot \frac{\theta}{2} + \tan \frac{\theta}{2} \right) = \frac{2r'}{\sin \theta}.$$
But we observe that
$$DP = \frac{BD + DA - AB}{2}, \quad DQ = \frac{BD + DC - BC}{2}.$$
Thus $PQ = (b - c - a + 2BD)/2$. We also have
$$\begin{aligned} \frac{ac}{2} &= [ABC] = [ABD] + [CBD] = r' \frac{(AB + BD + DA)}{2} + r' \frac{(CB + BD + DC)}{2} \\ &= r' \frac{(c+a+b+2BD)}{2} = r'(s+BD). \end{aligned}$$
But
$$r' = \frac{PQ \sin \theta}{2} = \frac{PQ \cdot h}{2BD},$$
where $h$ is the altitude from $B$ on to $AC$. But we know that $h = ac/b$. Thus we get
$$ac = 2 \times r'(s + BD) = 2 \times \frac{PQ \cdot h}{2 \times BD} (s + BD) = \frac{(b - c - a + 2BD)ca(s + BD)}{2 \times BD \times b}.$$
Thus we get
$$2 \times BD \times b = 2 \times (BD - (s - b))(s + BD).$$
This gives $BD^2 = s(s - b)$. Since $ABC$ is a right-angled triangle $r = s - b$. Thus we get $BD^2 = rs$. On the other hand, we also have $[ABC] = r'(s + BD)$. Thus we get
$$rs = [ABC] = r'(s + BD).$$
Hence
$$\frac{1}{r'} = \frac{1}{r} + \frac{BD}{rs} = \frac{1}{r} + \frac{1}{BD}.$$

Alternate Solution 1:

Observe that
$$\frac{r'}{r} = \frac{AP}{AX} = \frac{CQ}{CX} = \frac{AP + CQ}{AC},$$
where $X$ is the point at which the incircle of $ABC$ touches the side $AC$. If $s_1$ and $s_2$ are respectively the semi-perimeters of triangles $ABD$ and $CBD$, we know $AP = s_1 - BD$ and $CQ = s_2 - BD$. Therefore
$$\frac{r'}{r} = \frac{(s_1 - BD) + (s_2 - BD)}{b} = \frac{s_1 + s_2 - 2BD}{b}.$$
But
$$s_1 + s_2 = \frac{AD + BD + c}{2} + \frac{CD + BD + a}{2} = \frac{(a + b + c) + 2BD}{2} = \frac{s + BD}{2}.$$
This gives
$$\frac{r'}{r} = \frac{s + BD - 2BD}{b} = \frac{s - BD}{b}.$$
We also have
$$r' = \frac{[ABD]}{s_1} = \frac{[CBD]}{s_2} = \frac{[ABD] + [CBD]}{s_1 + s_2} = \frac{[ABC]}{s + BD} = \frac{rs}{s + BD}.$$
This implies that
$$\frac{r'}{r} = \frac{s}{s + BD}.$$
Comparing the two expressions for $r'/r$, we see that
$$\frac{s - BD}{b} = \frac{s}{s + BD}.$$
Therefore $s^2 - BD^2 = bs$, or $BD^2 = s(s-b)$. Thus we get $BD = \sqrt{s(s-b)}$.
We know now that
$$\frac{r'}{r} = \frac{s}{s + BD} = \frac{s - BD}{b} = \frac{BD}{(s-b) + BD} = \frac{\sqrt{s(s-b)}}{(s-b) + \sqrt{s(s-b)}} = \frac{\sqrt{s}}{\sqrt{s-b} + \sqrt{s}}.$$
Therefore
$$\frac{r}{r'} = 1 + \sqrt{\frac{s-b}{s}}.$$
This gives
$$\frac{1}{r'} = \frac{1}{r} + \left( \sqrt{\frac{s-b}{s}} \right) \frac{1}{r}.$$
But
$$\left( \sqrt{\frac{s-b}{s}} \right) \frac{1}{r} = \left( \frac{s-b}{\sqrt{s(s-b)}} \right) \frac{1}{r} = \left( \frac{s-b}{BD} \right) \frac{1}{r}.$$
If $\angle B = 90^\circ$, we know that $r = s - b$. Therefore we get
$$\frac{1}{r'} = \frac{1}{r} + \left( \frac{s-b}{BD} \right) \frac{1}{r} = \frac{1}{r} + \frac{1}{BD}.$$

Alternate Solution 2:

Observe that $\angle EDF = 90^\circ$. Hence $\triangle EDP$ is similar to $\triangle DFQ$. Therefore $DP \cdot DQ = EP \cdot FQ$. Taking $DP = x_2$ and $DQ = x_1$, we get $x_1x_2 = (r')^2$. We also observe that $BD = x_2 + y_2 = x_1 + y_1$. Since $\angle EBF = 45^\circ$, we get

$$1 = \tan 45^\circ = \tan(\beta_1 + \beta_2) = \frac{\tan \beta_1 + \tan \beta_2}{1 - \tan \beta_1 \tan \beta_2}.$$
But $\tan \beta_1 = r'/y_2$ and $\tan \beta_2 = r'/x_2$. Hence we obtain
$$1 = \frac{(r'/y_2) + (r'/x_2)}{1 - (r')^2/x_2y_2}.$$
Solving for $r'$, we get
$$r' = \frac{x_2 y_2 - x_1 y_1}{x_2 + y_2}.$$
We also know
$$r = \frac{AB + BC - AC}{2} = \frac{x_2 + y_2 - (x_1 + y_1)}{2} = \frac{(x_2 - x_1) + (y_2 - y_1)}{2}.$$
Finally,
$$\begin{aligned} \frac{1}{r} + \frac{1}{BD} &= \frac{2}{(x_2 - x_1) + (y_2 - y_1)} + \frac{1}{x_1 + x_2} \\ &= \frac{2x_1 + 2x_2 + (x_2 - x_1) + (y_2 - y_1)}{(x_1 + x_2)((x_2 - x_1) + (y_2 - y_1))}. \end{aligned}$$
But we can write
$$2x_1 + 2x_2 + (x_2 - x_1) + (y_2 - y_1) = (x_1 + x_2 + x_2 - x_1) + (y_1 + y_2 + y_2 - y_1) = 2(x_2 + y_2),$$
and
$$(x_1 + x_2)((x_2 - x_1) + (y_2 - y_1)) = 2(x_1 + x_2)(x_2 - y_1) \\ = 2(x_2(x_2 + x_1 - y_1) - x_1y_1) = 2(x_2y_2 - x_1y_1).$$
Therefore
$$\frac{1}{r} + \frac{1}{BD} = \frac{2(x_2 + y_2)}{2(x_2y_2 - x_1y_1)} = \frac{1}{r'}.$$